We consider the system Applicative_first_order_05__#3.10.

  Alphabet:

    0 : [] --> b 
    add : [b * c] --> c 
    app : [c * c] --> c 
    eq : [b * b] --> a 
    false : [] --> a 
    filter : [b -> a * c] --> c 
    filter2 : [a * b -> a * b * c] --> c 
    if!fac6220min : [a * c] --> b 
    if!fac6220minsort : [a * c * c] --> c 
    if!fac6220rm : [a * b * c] --> c 
    le : [b * b] --> a 
    map : [b -> b * c] --> c 
    min : [c] --> b 
    minsort : [c * c] --> c 
    nil : [] --> c 
    rm : [b * c] --> c 
    s : [b] --> b 
    true : [] --> a 

  Rules:

    eq(0, 0) => true 
    eq(0, s(x)) => false 
    eq(s(x), 0) => false 
    eq(s(x), s(y)) => eq(x, y) 
    le(0, x) => true 
    le(s(x), 0) => false 
    le(s(x), s(y)) => le(x, y) 
    app(nil, x) => x 
    app(add(x, y), z) => add(x, app(y, z)) 
    min(add(x, nil)) => x 
    min(add(x, add(y, z))) => if!fac6220min(le(x, y), add(x, add(y, z))) 
    if!fac6220min(true, add(x, add(y, z))) => min(add(x, z)) 
    if!fac6220min(false, add(x, add(y, z))) => min(add(y, z)) 
    rm(x, nil) => nil 
    rm(x, add(y, z)) => if!fac6220rm(eq(x, y), x, add(y, z)) 
    if!fac6220rm(true, x, add(y, z)) => rm(x, z) 
    if!fac6220rm(false, x, add(y, z)) => add(y, rm(x, z)) 
    minsort(nil, nil) => nil 
    minsort(add(x, y), z) => if!fac6220minsort(eq(x, min(add(x, y))), add(x, y), z) 
    if!fac6220minsort(true, add(x, y), z) => add(x, minsort(app(rm(x, y), z), nil)) 
    if!fac6220minsort(false, add(x, y), z) => minsort(y, add(x, z)) 
    map(f, nil) => nil 
    map(f, add(x, y)) => add(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, add(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => add(x, filter(f, y)) 
    filter2(false, f, x, y) => filter(f, y) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We observe that the rules contain a first-order subset:

  eq(0, 0) => true 
  eq(0, s(X)) => false 
  eq(s(X), 0) => false 
  eq(s(X), s(Y)) => eq(X, Y) 
  le(0, X) => true 
  le(s(X), 0) => false 
  le(s(X), s(Y)) => le(X, Y) 
  app(nil, X) => X 
  app(add(X, Y), Z) => add(X, app(Y, Z)) 
  min(add(X, nil)) => X 
  min(add(X, add(Y, Z))) => if!fac6220min(le(X, Y), add(X, add(Y, Z))) 
  if!fac6220min(true, add(X, add(Y, Z))) => min(add(X, Z)) 
  if!fac6220min(false, add(X, add(Y, Z))) => min(add(Y, Z)) 
  rm(X, nil) => nil 
  rm(X, add(Y, Z)) => if!fac6220rm(eq(X, Y), X, add(Y, Z)) 
  if!fac6220rm(true, X, add(Y, Z)) => rm(X, Z) 
  if!fac6220rm(false, X, add(Y, Z)) => add(Y, rm(X, Z)) 
  minsort(nil, nil) => nil 
  minsort(add(X, Y), Z) => if!fac6220minsort(eq(X, min(add(X, Y))), add(X, Y), Z) 
  if!fac6220minsort(true, add(X, Y), Z) => add(X, minsort(app(rm(X, Y), Z), nil)) 
  if!fac6220minsort(false, add(X, Y), Z) => minsort(Y, add(X, Z)) 

Moreover, the system is orthogonal.  Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is terminating when seen as a many-sorted first-order TRS.

According to the external first-order termination prover, this system is indeed terminating:

 || proof of resources/system.trs
 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
 || 
 || 
 || Termination w.r.t. Q of the given QTRS could be proven:
 || 
 || (0) QTRS
 || (1) Overlay + Local Confluence [EQUIVALENT]
 || (2) QTRS
 || (3) DependencyPairsProof [EQUIVALENT]
 || (4) QDP
 || (5) DependencyGraphProof [EQUIVALENT]
 || (6) AND
 ||     (7) QDP
 ||         (8) UsableRulesProof [EQUIVALENT]
 ||         (9) QDP
 ||         (10) QReductionProof [EQUIVALENT]
 ||         (11) QDP
 ||         (12) QDPSizeChangeProof [EQUIVALENT]
 ||         (13) YES
 ||     (14) QDP
 ||         (15) UsableRulesProof [EQUIVALENT]
 ||         (16) QDP
 ||         (17) QReductionProof [EQUIVALENT]
 ||         (18) QDP
 ||         (19) QDPSizeChangeProof [EQUIVALENT]
 ||         (20) YES
 ||     (21) QDP
 ||         (22) UsableRulesProof [EQUIVALENT]
 ||         (23) QDP
 ||         (24) QReductionProof [EQUIVALENT]
 ||         (25) QDP
 ||         (26) QDPOrderProof [EQUIVALENT]
 ||         (27) QDP
 ||         (28) DependencyGraphProof [EQUIVALENT]
 ||         (29) TRUE
 ||     (30) QDP
 ||         (31) UsableRulesProof [EQUIVALENT]
 ||         (32) QDP
 ||         (33) QReductionProof [EQUIVALENT]
 ||         (34) QDP
 ||         (35) QDPSizeChangeProof [EQUIVALENT]
 ||         (36) YES
 ||     (37) QDP
 ||         (38) UsableRulesProof [EQUIVALENT]
 ||         (39) QDP
 ||         (40) QReductionProof [EQUIVALENT]
 ||         (41) QDP
 ||         (42) QDPSizeChangeProof [EQUIVALENT]
 ||         (43) YES
 ||     (44) QDP
 ||         (45) UsableRulesProof [EQUIVALENT]
 ||         (46) QDP
 ||         (47) QReductionProof [EQUIVALENT]
 ||         (48) QDP
 ||         (49) QDPOrderProof [EQUIVALENT]
 ||         (50) QDP
 ||         (51) UsableRulesProof [EQUIVALENT]
 ||         (52) QDP
 ||         (53) QReductionProof [EQUIVALENT]
 ||         (54) QDP
 ||         (55) QDPSizeChangeProof [EQUIVALENT]
 ||         (56) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    rm(%X, nil) -> nil
 ||    rm(%X, add(%Y, %Z)) -> if!fac6220rm(eq(%X, %Y), %X, add(%Y, %Z))
 ||    if!fac6220rm(true, %X, add(%Y, %Z)) -> rm(%X, %Z)
 ||    if!fac6220rm(false, %X, add(%Y, %Z)) -> add(%Y, rm(%X, %Z))
 ||    minsort(nil, nil) -> nil
 ||    minsort(add(%X, %Y), %Z) -> if!fac6220minsort(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    if!fac6220minsort(true, add(%X, %Y), %Z) -> add(%X, minsort(app(rm(%X, %Y), %Z), nil))
 ||    if!fac6220minsort(false, add(%X, %Y), %Z) -> minsort(%Y, add(%X, %Z))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) Overlay + Local Confluence (EQUIVALENT)
 || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    rm(%X, nil) -> nil
 ||    rm(%X, add(%Y, %Z)) -> if!fac6220rm(eq(%X, %Y), %X, add(%Y, %Z))
 ||    if!fac6220rm(true, %X, add(%Y, %Z)) -> rm(%X, %Z)
 ||    if!fac6220rm(false, %X, add(%Y, %Z)) -> add(%Y, rm(%X, %Z))
 ||    minsort(nil, nil) -> nil
 ||    minsort(add(%X, %Y), %Z) -> if!fac6220minsort(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    if!fac6220minsort(true, add(%X, %Y), %Z) -> add(%X, minsort(app(rm(%X, %Y), %Z), nil))
 ||    if!fac6220minsort(false, add(%X, %Y), %Z) -> minsort(%Y, add(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || 
 || ----------------------------------------
 || 
 || (3) DependencyPairsProof (EQUIVALENT)
 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
 || ----------------------------------------
 || 
 || (4)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 ||    APP(add(%X, %Y), %Z) -> APP(%Y, %Z)
 ||    MIN(add(%X, add(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    MIN(add(%X, add(%Y, %Z))) -> LE(%X, %Y)
 ||    IF!FAC6220MIN(true, add(%X, add(%Y, %Z))) -> MIN(add(%X, %Z))
 ||    IF!FAC6220MIN(false, add(%X, add(%Y, %Z))) -> MIN(add(%Y, %Z))
 ||    RM(%X, add(%Y, %Z)) -> IF!FAC6220RM(eq(%X, %Y), %X, add(%Y, %Z))
 ||    RM(%X, add(%Y, %Z)) -> EQ(%X, %Y)
 ||    IF!FAC6220RM(true, %X, add(%Y, %Z)) -> RM(%X, %Z)
 ||    IF!FAC6220RM(false, %X, add(%Y, %Z)) -> RM(%X, %Z)
 ||    MINSORT(add(%X, %Y), %Z) -> IF!FAC6220MINSORT(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    MINSORT(add(%X, %Y), %Z) -> EQ(%X, min(add(%X, %Y)))
 ||    MINSORT(add(%X, %Y), %Z) -> MIN(add(%X, %Y))
 ||    IF!FAC6220MINSORT(true, add(%X, %Y), %Z) -> MINSORT(app(rm(%X, %Y), %Z), nil)
 ||    IF!FAC6220MINSORT(true, add(%X, %Y), %Z) -> APP(rm(%X, %Y), %Z)
 ||    IF!FAC6220MINSORT(true, add(%X, %Y), %Z) -> RM(%X, %Y)
 ||    IF!FAC6220MINSORT(false, add(%X, %Y), %Z) -> MINSORT(%Y, add(%X, %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    rm(%X, nil) -> nil
 ||    rm(%X, add(%Y, %Z)) -> if!fac6220rm(eq(%X, %Y), %X, add(%Y, %Z))
 ||    if!fac6220rm(true, %X, add(%Y, %Z)) -> rm(%X, %Z)
 ||    if!fac6220rm(false, %X, add(%Y, %Z)) -> add(%Y, rm(%X, %Z))
 ||    minsort(nil, nil) -> nil
 ||    minsort(add(%X, %Y), %Z) -> if!fac6220minsort(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    if!fac6220minsort(true, add(%X, %Y), %Z) -> add(%X, minsort(app(rm(%X, %Y), %Z), nil))
 ||    if!fac6220minsort(false, add(%X, %Y), %Z) -> minsort(%Y, add(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (5) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 6 less nodes.
 || ----------------------------------------
 || 
 || (6)
 || Complex Obligation (AND)
 || 
 || ----------------------------------------
 || 
 || (7)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    APP(add(%X, %Y), %Z) -> APP(%Y, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    rm(%X, nil) -> nil
 ||    rm(%X, add(%Y, %Z)) -> if!fac6220rm(eq(%X, %Y), %X, add(%Y, %Z))
 ||    if!fac6220rm(true, %X, add(%Y, %Z)) -> rm(%X, %Z)
 ||    if!fac6220rm(false, %X, add(%Y, %Z)) -> add(%Y, rm(%X, %Z))
 ||    minsort(nil, nil) -> nil
 ||    minsort(add(%X, %Y), %Z) -> if!fac6220minsort(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    if!fac6220minsort(true, add(%X, %Y), %Z) -> add(%X, minsort(app(rm(%X, %Y), %Z), nil))
 ||    if!fac6220minsort(false, add(%X, %Y), %Z) -> minsort(%Y, add(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (8) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (9)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    APP(add(%X, %Y), %Z) -> APP(%Y, %Z)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (10) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || 
 || ----------------------------------------
 || 
 || (11)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    APP(add(%X, %Y), %Z) -> APP(%Y, %Z)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (12) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *APP(add(%X, %Y), %Z) -> APP(%Y, %Z)
 || The graph contains the following edges 1 > 1, 2 >= 2
 || 
 || 
 || ----------------------------------------
 || 
 || (13)
 || YES
 || 
 || ----------------------------------------
 || 
 || (14)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    rm(%X, nil) -> nil
 ||    rm(%X, add(%Y, %Z)) -> if!fac6220rm(eq(%X, %Y), %X, add(%Y, %Z))
 ||    if!fac6220rm(true, %X, add(%Y, %Z)) -> rm(%X, %Z)
 ||    if!fac6220rm(false, %X, add(%Y, %Z)) -> add(%Y, rm(%X, %Z))
 ||    minsort(nil, nil) -> nil
 ||    minsort(add(%X, %Y), %Z) -> if!fac6220minsort(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    if!fac6220minsort(true, add(%X, %Y), %Z) -> add(%X, minsort(app(rm(%X, %Y), %Z), nil))
 ||    if!fac6220minsort(false, add(%X, %Y), %Z) -> minsort(%Y, add(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (15) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (16)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (17) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || 
 || ----------------------------------------
 || 
 || (18)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (19) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *LE(s(%X), s(%Y)) -> LE(%X, %Y)
 || The graph contains the following edges 1 > 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (20)
 || YES
 || 
 || ----------------------------------------
 || 
 || (21)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MIN(add(%X, add(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    IF!FAC6220MIN(true, add(%X, add(%Y, %Z))) -> MIN(add(%X, %Z))
 ||    IF!FAC6220MIN(false, add(%X, add(%Y, %Z))) -> MIN(add(%Y, %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    rm(%X, nil) -> nil
 ||    rm(%X, add(%Y, %Z)) -> if!fac6220rm(eq(%X, %Y), %X, add(%Y, %Z))
 ||    if!fac6220rm(true, %X, add(%Y, %Z)) -> rm(%X, %Z)
 ||    if!fac6220rm(false, %X, add(%Y, %Z)) -> add(%Y, rm(%X, %Z))
 ||    minsort(nil, nil) -> nil
 ||    minsort(add(%X, %Y), %Z) -> if!fac6220minsort(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    if!fac6220minsort(true, add(%X, %Y), %Z) -> add(%X, minsort(app(rm(%X, %Y), %Z), nil))
 ||    if!fac6220minsort(false, add(%X, %Y), %Z) -> minsort(%Y, add(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (22) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (23)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MIN(add(%X, add(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    IF!FAC6220MIN(true, add(%X, add(%Y, %Z))) -> MIN(add(%X, %Z))
 ||    IF!FAC6220MIN(false, add(%X, add(%Y, %Z))) -> MIN(add(%Y, %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (24) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || 
 || ----------------------------------------
 || 
 || (25)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MIN(add(%X, add(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    IF!FAC6220MIN(true, add(%X, add(%Y, %Z))) -> MIN(add(%X, %Z))
 ||    IF!FAC6220MIN(false, add(%X, add(%Y, %Z))) -> MIN(add(%Y, %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (26) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    IF!FAC6220MIN(true, add(%X, add(%Y, %Z))) -> MIN(add(%X, %Z))
 ||    IF!FAC6220MIN(false, add(%X, add(%Y, %Z))) -> MIN(add(%Y, %Z))
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
 || 
 || POL( IF!FAC6220MIN_2(x_1, x_2) ) = max{0, 2x_1 + 2x_2 - 2}
 || POL( le_2(x_1, x_2) ) = 2
 || POL( 0 ) = 2
 || POL( true ) = 0
 || POL( s_1(x_1) ) = 2
 || POL( false ) = 0
 || POL( MIN_1(x_1) ) = 2x_1 + 2
 || POL( add_2(x_1, x_2) ) = x_1 + 2x_2 + 2
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || 
 || ----------------------------------------
 || 
 || (27)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MIN(add(%X, add(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), add(%X, add(%Y, %Z)))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (28) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
 || ----------------------------------------
 || 
 || (29)
 || TRUE
 || 
 || ----------------------------------------
 || 
 || (30)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    rm(%X, nil) -> nil
 ||    rm(%X, add(%Y, %Z)) -> if!fac6220rm(eq(%X, %Y), %X, add(%Y, %Z))
 ||    if!fac6220rm(true, %X, add(%Y, %Z)) -> rm(%X, %Z)
 ||    if!fac6220rm(false, %X, add(%Y, %Z)) -> add(%Y, rm(%X, %Z))
 ||    minsort(nil, nil) -> nil
 ||    minsort(add(%X, %Y), %Z) -> if!fac6220minsort(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    if!fac6220minsort(true, add(%X, %Y), %Z) -> add(%X, minsort(app(rm(%X, %Y), %Z), nil))
 ||    if!fac6220minsort(false, add(%X, %Y), %Z) -> minsort(%Y, add(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (31) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (32)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (33) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || 
 || ----------------------------------------
 || 
 || (34)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (35) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *EQ(s(%X), s(%Y)) -> EQ(%X, %Y)
 || The graph contains the following edges 1 > 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (36)
 || YES
 || 
 || ----------------------------------------
 || 
 || (37)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    RM(%X, add(%Y, %Z)) -> IF!FAC6220RM(eq(%X, %Y), %X, add(%Y, %Z))
 ||    IF!FAC6220RM(true, %X, add(%Y, %Z)) -> RM(%X, %Z)
 ||    IF!FAC6220RM(false, %X, add(%Y, %Z)) -> RM(%X, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    rm(%X, nil) -> nil
 ||    rm(%X, add(%Y, %Z)) -> if!fac6220rm(eq(%X, %Y), %X, add(%Y, %Z))
 ||    if!fac6220rm(true, %X, add(%Y, %Z)) -> rm(%X, %Z)
 ||    if!fac6220rm(false, %X, add(%Y, %Z)) -> add(%Y, rm(%X, %Z))
 ||    minsort(nil, nil) -> nil
 ||    minsort(add(%X, %Y), %Z) -> if!fac6220minsort(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    if!fac6220minsort(true, add(%X, %Y), %Z) -> add(%X, minsort(app(rm(%X, %Y), %Z), nil))
 ||    if!fac6220minsort(false, add(%X, %Y), %Z) -> minsort(%Y, add(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (38) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (39)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    RM(%X, add(%Y, %Z)) -> IF!FAC6220RM(eq(%X, %Y), %X, add(%Y, %Z))
 ||    IF!FAC6220RM(true, %X, add(%Y, %Z)) -> RM(%X, %Z)
 ||    IF!FAC6220RM(false, %X, add(%Y, %Z)) -> RM(%X, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (40) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || 
 || ----------------------------------------
 || 
 || (41)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    RM(%X, add(%Y, %Z)) -> IF!FAC6220RM(eq(%X, %Y), %X, add(%Y, %Z))
 ||    IF!FAC6220RM(true, %X, add(%Y, %Z)) -> RM(%X, %Z)
 ||    IF!FAC6220RM(false, %X, add(%Y, %Z)) -> RM(%X, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (42) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *RM(%X, add(%Y, %Z)) -> IF!FAC6220RM(eq(%X, %Y), %X, add(%Y, %Z))
 || The graph contains the following edges 1 >= 2, 2 >= 3
 || 
 || 
 || *IF!FAC6220RM(true, %X, add(%Y, %Z)) -> RM(%X, %Z)
 || The graph contains the following edges 2 >= 1, 3 > 2
 || 
 || 
 || *IF!FAC6220RM(false, %X, add(%Y, %Z)) -> RM(%X, %Z)
 || The graph contains the following edges 2 >= 1, 3 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (43)
 || YES
 || 
 || ----------------------------------------
 || 
 || (44)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    IF!FAC6220MINSORT(true, add(%X, %Y), %Z) -> MINSORT(app(rm(%X, %Y), %Z), nil)
 ||    MINSORT(add(%X, %Y), %Z) -> IF!FAC6220MINSORT(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    IF!FAC6220MINSORT(false, add(%X, %Y), %Z) -> MINSORT(%Y, add(%X, %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    rm(%X, nil) -> nil
 ||    rm(%X, add(%Y, %Z)) -> if!fac6220rm(eq(%X, %Y), %X, add(%Y, %Z))
 ||    if!fac6220rm(true, %X, add(%Y, %Z)) -> rm(%X, %Z)
 ||    if!fac6220rm(false, %X, add(%Y, %Z)) -> add(%Y, rm(%X, %Z))
 ||    minsort(nil, nil) -> nil
 ||    minsort(add(%X, %Y), %Z) -> if!fac6220minsort(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    if!fac6220minsort(true, add(%X, %Y), %Z) -> add(%X, minsort(app(rm(%X, %Y), %Z), nil))
 ||    if!fac6220minsort(false, add(%X, %Y), %Z) -> minsort(%Y, add(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (45) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (46)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    IF!FAC6220MINSORT(true, add(%X, %Y), %Z) -> MINSORT(app(rm(%X, %Y), %Z), nil)
 ||    MINSORT(add(%X, %Y), %Z) -> IF!FAC6220MINSORT(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    IF!FAC6220MINSORT(false, add(%X, %Y), %Z) -> MINSORT(%Y, add(%X, %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    rm(%X, nil) -> nil
 ||    rm(%X, add(%Y, %Z)) -> if!fac6220rm(eq(%X, %Y), %X, add(%Y, %Z))
 ||    if!fac6220rm(true, %X, add(%Y, %Z)) -> rm(%X, %Z)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    if!fac6220rm(false, %X, add(%Y, %Z)) -> add(%Y, rm(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (47) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    minsort(nil, nil)
 ||    minsort(add(x0, x1), x2)
 ||    if!fac6220minsort(true, add(x0, x1), x2)
 ||    if!fac6220minsort(false, add(x0, x1), x2)
 || 
 || 
 || ----------------------------------------
 || 
 || (48)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    IF!FAC6220MINSORT(true, add(%X, %Y), %Z) -> MINSORT(app(rm(%X, %Y), %Z), nil)
 ||    MINSORT(add(%X, %Y), %Z) -> IF!FAC6220MINSORT(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    IF!FAC6220MINSORT(false, add(%X, %Y), %Z) -> MINSORT(%Y, add(%X, %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    rm(%X, nil) -> nil
 ||    rm(%X, add(%Y, %Z)) -> if!fac6220rm(eq(%X, %Y), %X, add(%Y, %Z))
 ||    if!fac6220rm(true, %X, add(%Y, %Z)) -> rm(%X, %Z)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    if!fac6220rm(false, %X, add(%Y, %Z)) -> add(%Y, rm(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (49) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    IF!FAC6220MINSORT(true, add(%X, %Y), %Z) -> MINSORT(app(rm(%X, %Y), %Z), nil)
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 0
 ||    POL(IF!FAC6220MINSORT(x_1, x_2, x_3)) = x_2 + x_3
 ||    POL(MINSORT(x_1, x_2)) = x_1 + x_2
 ||    POL(add(x_1, x_2)) = 1 + x_2
 ||    POL(app(x_1, x_2)) = x_1 + x_2
 ||    POL(eq(x_1, x_2)) = 0
 ||    POL(false) = 0
 ||    POL(if!fac6220min(x_1, x_2)) = 0
 ||    POL(if!fac6220rm(x_1, x_2, x_3)) = x_3
 ||    POL(le(x_1, x_2)) = 0
 ||    POL(min(x_1)) = 0
 ||    POL(nil) = 0
 ||    POL(rm(x_1, x_2)) = x_2
 ||    POL(s(x_1)) = 0
 ||    POL(true) = 0
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    rm(%X, nil) -> nil
 ||    rm(%X, add(%Y, %Z)) -> if!fac6220rm(eq(%X, %Y), %X, add(%Y, %Z))
 ||    if!fac6220rm(true, %X, add(%Y, %Z)) -> rm(%X, %Z)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    if!fac6220rm(false, %X, add(%Y, %Z)) -> add(%Y, rm(%X, %Z))
 || 
 || 
 || ----------------------------------------
 || 
 || (50)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINSORT(add(%X, %Y), %Z) -> IF!FAC6220MINSORT(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    IF!FAC6220MINSORT(false, add(%X, %Y), %Z) -> MINSORT(%Y, add(%X, %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    rm(%X, nil) -> nil
 ||    rm(%X, add(%Y, %Z)) -> if!fac6220rm(eq(%X, %Y), %X, add(%Y, %Z))
 ||    if!fac6220rm(true, %X, add(%Y, %Z)) -> rm(%X, %Z)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    if!fac6220rm(false, %X, add(%Y, %Z)) -> add(%Y, rm(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (51) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (52)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINSORT(add(%X, %Y), %Z) -> IF!FAC6220MINSORT(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    IF!FAC6220MINSORT(false, add(%X, %Y), %Z) -> MINSORT(%Y, add(%X, %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (53) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    rm(x0, nil)
 ||    rm(x0, add(x1, x2))
 ||    if!fac6220rm(true, x0, add(x1, x2))
 ||    if!fac6220rm(false, x0, add(x1, x2))
 || 
 || 
 || ----------------------------------------
 || 
 || (54)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINSORT(add(%X, %Y), %Z) -> IF!FAC6220MINSORT(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 ||    IF!FAC6220MINSORT(false, add(%X, %Y), %Z) -> MINSORT(%Y, add(%X, %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    min(add(%X, nil)) -> %X
 ||    min(add(%X, add(%Y, %Z))) -> if!fac6220min(le(%X, %Y), add(%X, add(%Y, %Z)))
 ||    if!fac6220min(true, add(%X, add(%Y, %Z))) -> min(add(%X, %Z))
 ||    if!fac6220min(false, add(%X, add(%Y, %Z))) -> min(add(%Y, %Z))
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    eq(0, 0)
 ||    eq(0, s(x0))
 ||    eq(s(x0), 0)
 ||    eq(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(add(x0, nil))
 ||    min(add(x0, add(x1, x2)))
 ||    if!fac6220min(true, add(x0, add(x1, x2)))
 ||    if!fac6220min(false, add(x0, add(x1, x2)))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (55) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *IF!FAC6220MINSORT(false, add(%X, %Y), %Z) -> MINSORT(%Y, add(%X, %Z))
 || The graph contains the following edges 2 > 1
 || 
 || 
 || *MINSORT(add(%X, %Y), %Z) -> IF!FAC6220MINSORT(eq(%X, min(add(%X, %Y))), add(%X, %Y), %Z)
 || The graph contains the following edges 1 >= 2, 2 >= 3
 || 
 || 
 || ----------------------------------------
 || 
 || (56)
 || YES
 || 
We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.

After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] map#(F, add(X, Y)) =#> F(X)   
    1] map#(F, add(X, Y)) =#> map#(F, Y)   
    2] filter#(F, add(X, Y)) =#> filter2#(F X, F, X, Y)   
    3] filter#(F, add(X, Y)) =#> F(X)   
    4] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    5] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    eq(0, 0) => true 
    eq(0, s(X)) => false 
    eq(s(X), 0) => false 
    eq(s(X), s(Y)) => eq(X, Y) 
    le(0, X) => true 
    le(s(X), 0) => false 
    le(s(X), s(Y)) => le(X, Y) 
    app(nil, X) => X 
    app(add(X, Y), Z) => add(X, app(Y, Z)) 
    min(add(X, nil)) => X 
    min(add(X, add(Y, Z))) => if!fac6220min(le(X, Y), add(X, add(Y, Z))) 
    if!fac6220min(true, add(X, add(Y, Z))) => min(add(X, Z)) 
    if!fac6220min(false, add(X, add(Y, Z))) => min(add(Y, Z)) 
    rm(X, nil) => nil 
    rm(X, add(Y, Z)) => if!fac6220rm(eq(X, Y), X, add(Y, Z)) 
    if!fac6220rm(true, X, add(Y, Z)) => rm(X, Z) 
    if!fac6220rm(false, X, add(Y, Z)) => add(Y, rm(X, Z)) 
    minsort(nil, nil) => nil 
    minsort(add(X, Y), Z) => if!fac6220minsort(eq(X, min(add(X, Y))), add(X, Y), Z) 
    if!fac6220minsort(true, add(X, Y), Z) => add(X, minsort(app(rm(X, Y), Z), nil)) 
    if!fac6220minsort(false, add(X, Y), Z) => minsort(Y, add(X, Z)) 
    map(F, nil) => nil 
    map(F, add(X, Y)) => add(F X, map(F, Y)) 
    filter(F, nil) => nil 
    filter(F, add(X, Y)) => filter2(F X, F, X, Y) 
    filter2(true, F, X, Y) => add(X, filter(F, Y)) 
    filter2(false, F, X, Y) => filter(F, Y) 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70].  Thus, we must orient:

  map#(F, add(X, Y)) >? F(X) 
  map#(F, add(X, Y)) >? map#(F, Y) 
  filter#(F, add(X, Y)) >? filter2#(F X, F, X, Y) 
  filter#(F, add(X, Y)) >? F(X) 
  filter2#(true, F, X, Y) >? filter#(F, Y) 
  filter2#(false, F, X, Y) >? filter#(F, Y) 
  eq(0, 0) >= true 
  eq(0, s(X)) >= false 
  eq(s(X), 0) >= false 
  eq(s(X), s(Y)) >= eq(X, Y) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  app(nil, X) >= X 
  app(add(X, Y), Z) >= add(X, app(Y, Z)) 
  min(add(X, nil)) >= X 
  min(add(X, add(Y, Z))) >= if!fac6220min(le(X, Y), add(X, add(Y, Z))) 
  if!fac6220min(true, add(X, add(Y, Z))) >= min(add(X, Z)) 
  if!fac6220min(false, add(X, add(Y, Z))) >= min(add(Y, Z)) 
  rm(X, nil) >= nil 
  rm(X, add(Y, Z)) >= if!fac6220rm(eq(X, Y), X, add(Y, Z)) 
  if!fac6220rm(true, X, add(Y, Z)) >= rm(X, Z) 
  if!fac6220rm(false, X, add(Y, Z)) >= add(Y, rm(X, Z)) 
  minsort(nil, nil) >= nil 
  minsort(add(X, Y), Z) >= if!fac6220minsort(eq(X, min(add(X, Y))), add(X, Y), Z) 
  if!fac6220minsort(true, add(X, Y), Z) >= add(X, minsort(app(rm(X, Y), Z), nil)) 
  if!fac6220minsort(false, add(X, Y), Z) >= minsort(Y, add(X, Z)) 
  map(F, nil) >= nil 
  map(F, add(X, Y)) >= add(F X, map(F, Y)) 
  filter(F, nil) >= nil 
  filter(F, add(X, Y)) >= filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) >= add(X, filter(F, Y)) 
  filter2(false, F, X, Y) >= filter(F, Y) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 3 
  add = \y0y1.1 + y1 + 2y0 
  app = \y0y1.y0 + y1 
  eq = \y0y1.2 
  false = 0 
  filter = \G0y1.y1 + G0(0) + G0(y1) + 2y1G0(y1) 
  filter2 = \y0G1y2y3.1 + y3 + 2y2 + G1(0) + G1(y3) + 2y3G1(y3) 
  filter2# = \y0G1y2y3.1 + y3G1(y3) 
  filter# = \G0y1.1 + y1G0(y1) 
  if!fac6220min = \y0y1.2y1 
  if!fac6220minsort = \y0y1y2.2y1 + 2y2 
  if!fac6220rm = \y0y1y2.y1 + y2 
  le = \y0y1.y1 
  map = \G0y1.3 + y1 + 2y1G0(y1) + 2G0(y1) 
  map# = \G0y1.3 + G0(y1) 
  min = \y0.2y0 
  minsort = \y0y1.2y0 + 2y1 
  nil = 0 
  rm = \y0y1.y0 + y1 
  s = \y0.3 + 2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, add(_x1, _x2))]] = 3 + F0(1 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, add(_x1, _x2))]] = 3 + F0(1 + x2 + 2x1) >= 3 + F0(x2) = [[map#(_F0, _x2)]] 
  [[filter#(_F0, add(_x1, _x2))]] = 1 + F0(1 + x2 + 2x1) + 2x1F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) >= 1 + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter#(_F0, add(_x1, _x2))]] = 1 + F0(1 + x2 + 2x1) + 2x1F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 1 + x2F0(x2) >= 1 + x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 1 + x2F0(x2) >= 1 + x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[eq(0, 0)]] = 2 >= 0 = [[true]] 
  [[eq(0, s(_x0))]] = 2 >= 0 = [[false]] 
  [[eq(s(_x0), 0)]] = 2 >= 0 = [[false]] 
  [[eq(s(_x0), s(_x1))]] = 2 >= 2 = [[eq(_x0, _x1)]] 
  [[le(0, _x0)]] = x0 >= 0 = [[true]] 
  [[le(s(_x0), 0)]] = 3 >= 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 3 + 2x1 >= x1 = [[le(_x0, _x1)]] 
  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(add(_x0, _x1), _x2)]] = 1 + x1 + x2 + 2x0 >= 1 + x1 + x2 + 2x0 = [[add(_x0, app(_x1, _x2))]] 
  [[min(add(_x0, nil))]] = 2 + 4x0 >= x0 = [[_x0]] 
  [[min(add(_x0, add(_x1, _x2)))]] = 4 + 2x2 + 4x0 + 4x1 >= 4 + 2x2 + 4x0 + 4x1 = [[if!fac6220min(le(_x0, _x1), add(_x0, add(_x1, _x2)))]] 
  [[if!fac6220min(true, add(_x0, add(_x1, _x2)))]] = 4 + 2x2 + 4x0 + 4x1 >= 2 + 2x2 + 4x0 = [[min(add(_x0, _x2))]] 
  [[if!fac6220min(false, add(_x0, add(_x1, _x2)))]] = 4 + 2x2 + 4x0 + 4x1 >= 2 + 2x2 + 4x1 = [[min(add(_x1, _x2))]] 
  [[rm(_x0, nil)]] = x0 >= 0 = [[nil]] 
  [[rm(_x0, add(_x1, _x2))]] = 1 + x0 + x2 + 2x1 >= 1 + x0 + x2 + 2x1 = [[if!fac6220rm(eq(_x0, _x1), _x0, add(_x1, _x2))]] 
  [[if!fac6220rm(true, _x0, add(_x1, _x2))]] = 1 + x0 + x2 + 2x1 >= x0 + x2 = [[rm(_x0, _x2)]] 
  [[if!fac6220rm(false, _x0, add(_x1, _x2))]] = 1 + x0 + x2 + 2x1 >= 1 + x0 + x2 + 2x1 = [[add(_x1, rm(_x0, _x2))]] 
  [[minsort(nil, nil)]] = 0 >= 0 = [[nil]] 
  [[minsort(add(_x0, _x1), _x2)]] = 2 + 2x1 + 2x2 + 4x0 >= 2 + 2x1 + 2x2 + 4x0 = [[if!fac6220minsort(eq(_x0, min(add(_x0, _x1))), add(_x0, _x1), _x2)]] 
  [[if!fac6220minsort(true, add(_x0, _x1), _x2)]] = 2 + 2x1 + 2x2 + 4x0 >= 1 + 2x1 + 2x2 + 4x0 = [[add(_x0, minsort(app(rm(_x0, _x1), _x2), nil))]] 
  [[if!fac6220minsort(false, add(_x0, _x1), _x2)]] = 2 + 2x1 + 2x2 + 4x0 >= 2 + 2x1 + 2x2 + 4x0 = [[minsort(_x1, add(_x0, _x2))]] 
  [[map(_F0, nil)]] = 3 + 2F0(0) >= 0 = [[nil]] 
  [[map(_F0, add(_x1, _x2))]] = 4 + x2 + 2x1 + 2x2F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) + 4F0(1 + x2 + 2x1) >= 4 + x2 + 2x2F0(x2) + 2F0(x2) + 2max(x1, F0(x1)) = [[add(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] 
  [[filter(_F0, add(_x1, _x2))]] = 1 + x2 + 2x1 + F0(0) + 2x2F0(1 + x2 + 2x1) + 3F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) >= 1 + x2 + 2x1 + F0(0) + F0(x2) + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 1 + x2 + 2x1 + F0(0) + F0(x2) + 2x2F0(x2) >= 1 + x2 + 2x1 + F0(0) + F0(x2) + 2x2F0(x2) = [[add(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 1 + x2 + 2x1 + F0(0) + F0(x2) + 2x2F0(x2) >= x2 + F0(0) + F0(x2) + 2x2F0(x2) = [[filter(_F0, _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_1, R_0, minimal, formative), where P_1 consists of:

  map#(F, add(X, Y)) =#> map#(F, Y)   
  filter#(F, add(X, Y)) =#> filter2#(F X, F, X, Y)   
  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, Y)   

Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_0, minimal, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 0 
    * 1 : 2, 3 
    * 2 : 1 
    * 3 : 1 

This graph has the following strongly connected components:

  P_2:

    map#(F, add(X, Y)) =#> map#(F, Y)   

  P_3:

    filter#(F, add(X, Y)) =#> filter2#(F X, F, X, Y)   
    filter2#(true, F, X, Y) =#> filter#(F, Y)   
    filter2#(false, F, X, Y) =#> filter#(F, Y)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_1, R_0, m, f) by (P_2, R_0, m, f) and (P_3, R_0, m, f).

Thus, the original system is terminating if each of (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(filter2#) = 4 
  nu(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(filter#(F, add(X, Y))) = add(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 
  nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
  nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_3, R_0, minimal, f) by (P_4, R_0, minimal, f), where P_4 contains:

  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, Y)   

Thus, the original system is terminating if each of (P_2, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_4, R_0, minimal, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  
    * 1 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

Thus, the original system is terminating if (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(map#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(map#(F, add(X, Y))) = add(X, Y) |> Y = nu(map#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[FuhKop19]  C. Fuhs, and C. Kop.  A static higher-order dependency pair framework.  In Proceedings of ESOP 2019, 2019.
[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.