We consider the system Applicative_AG01_innermost__#4.36.

  Alphabet:

    0 : [] --> b 
    cons : [b * 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!fac6220replace : [a * b * b * c] --> c 
    le : [b * b] --> a 
    map : [b -> b * c] --> c 
    min : [c] --> b 
    nil : [] --> c 
    replace : [b * b * c] --> c 
    s : [b] --> b 
    sort : [c] --> c 
    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) 
    min(cons(0, nil)) => 0 
    min(cons(s(x), nil)) => s(x) 
    min(cons(x, cons(y, z))) => if!fac6220min(le(x, y), cons(x, cons(y, z))) 
    if!fac6220min(true, cons(x, cons(y, z))) => min(cons(x, z)) 
    if!fac6220min(false, cons(x, cons(y, z))) => min(cons(y, z)) 
    replace(x, y, nil) => nil 
    replace(x, y, cons(z, u)) => if!fac6220replace(eq(x, z), x, y, cons(z, u)) 
    if!fac6220replace(true, x, y, cons(z, u)) => cons(y, u) 
    if!fac6220replace(false, x, y, cons(z, u)) => cons(z, replace(x, y, u)) 
    sort(nil) => nil 
    sort(cons(x, y)) => cons(min(cons(x, y)), sort(replace(min(cons(x, y)), x, y))) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => cons(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) 
  min(cons(0, nil)) => 0 
  min(cons(s(X), nil)) => s(X) 
  min(cons(X, cons(Y, Z))) => if!fac6220min(le(X, Y), cons(X, cons(Y, Z))) 
  if!fac6220min(true, cons(X, cons(Y, Z))) => min(cons(X, Z)) 
  if!fac6220min(false, cons(X, cons(Y, Z))) => min(cons(Y, Z)) 
  replace(X, Y, nil) => nil 
  replace(X, Y, cons(Z, U)) => if!fac6220replace(eq(X, Z), X, Y, cons(Z, U)) 
  if!fac6220replace(true, X, Y, cons(Z, U)) => cons(Y, U) 
  if!fac6220replace(false, X, Y, cons(Z, U)) => cons(Z, replace(X, Y, U)) 
  sort(nil) => nil 
  sort(cons(X, Y)) => cons(min(cons(X, Y)), sort(replace(min(cons(X, Y)), X, Y))) 

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) QDPOrderProof [EQUIVALENT]
 ||         (20) QDP
 ||         (21) DependencyGraphProof [EQUIVALENT]
 ||         (22) TRUE
 ||     (23) QDP
 ||         (24) UsableRulesProof [EQUIVALENT]
 ||         (25) QDP
 ||         (26) QReductionProof [EQUIVALENT]
 ||         (27) QDP
 ||         (28) QDPSizeChangeProof [EQUIVALENT]
 ||         (29) YES
 ||     (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) QDPOrderProof [EQUIVALENT]
 ||         (43) QDP
 ||         (44) PisEmptyProof [EQUIVALENT]
 ||         (45) 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)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %X, %Y)))
 || 
 || 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)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %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(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (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)
 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    MIN(cons(%X, cons(%Y, %Z))) -> LE(%X, %Y)
 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%Y, %Z))
 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> EQ(%X, %Z)
 ||    IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
 ||    SORT(cons(%X, %Y)) -> MIN(cons(%X, %Y))
 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
 ||    SORT(cons(%X, %Y)) -> REPLACE(min(cons(%X, %Y)), %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)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %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(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (5) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 4 less nodes.
 || ----------------------------------------
 || 
 || (6)
 || Complex Obligation (AND)
 || 
 || ----------------------------------------
 || 
 || (7)
 || 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)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %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(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 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:
 || 
 ||    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))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 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))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (11)
 || 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.
 || ----------------------------------------
 || 
 || (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:
 || *LE(s(%X), s(%Y)) -> LE(%X, %Y)
 || 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:
 || 
 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%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)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %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(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 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:
 || 
 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%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))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 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))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (18)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%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.
 || ----------------------------------------
 || 
 || (19) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    IF!FAC6220MIN(true, cons(%X, cons(%Y, %Z))) -> MIN(cons(%X, %Z))
 ||    IF!FAC6220MIN(false, cons(%X, cons(%Y, %Z))) -> MIN(cons(%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( cons_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)
 || 
 || 
 || ----------------------------------------
 || 
 || (20)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MIN(cons(%X, cons(%Y, %Z))) -> IF!FAC6220MIN(le(%X, %Y), cons(%X, cons(%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.
 || ----------------------------------------
 || 
 || (21) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
 || ----------------------------------------
 || 
 || (22)
 || TRUE
 || 
 || ----------------------------------------
 || 
 || (23)
 || 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)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %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(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (24) 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.
 || ----------------------------------------
 || 
 || (25)
 || 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))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (26) 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))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (27)
 || 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.
 || ----------------------------------------
 || 
 || (28) 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
 || 
 || 
 || ----------------------------------------
 || 
 || (29)
 || YES
 || 
 || ----------------------------------------
 || 
 || (30)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
 || 
 || 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)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %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(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 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:
 || 
 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
 || 
 || 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))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 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].
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    min(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (34)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
 || 
 || 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.
 || ----------------------------------------
 || 
 || (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:
 || *IF!FAC6220REPLACE(false, %X, %Y, cons(%Z, %U)) -> REPLACE(%X, %Y, %U)
 || The graph contains the following edges 2 >= 1, 3 >= 2, 4 > 3
 || 
 || 
 || *REPLACE(%X, %Y, cons(%Z, %U)) -> IF!FAC6220REPLACE(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 || The graph contains the following edges 1 >= 2, 2 >= 3, 3 >= 4
 || 
 || 
 || ----------------------------------------
 || 
 || (36)
 || YES
 || 
 || ----------------------------------------
 || 
 || (37)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %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)
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    sort(nil) -> nil
 ||    sort(cons(%X, %Y)) -> cons(min(cons(%X, %Y)), sort(replace(min(cons(%X, %Y)), %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(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 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:
 || 
 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    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(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 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].
 || 
 ||    sort(nil)
 ||    sort(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (41)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    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(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (42) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    SORT(cons(%X, %Y)) -> SORT(replace(min(cons(%X, %Y)), %X, %Y))
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 0
 ||    POL(SORT(x_1)) = x_1
 ||    POL(cons(x_1, x_2)) = 1 + x_2
 ||    POL(eq(x_1, x_2)) = 0
 ||    POL(false) = 0
 ||    POL(if!fac6220min(x_1, x_2)) = 0
 ||    POL(if!fac6220replace(x_1, x_2, x_3, x_4)) = x_4
 ||    POL(le(x_1, x_2)) = 0
 ||    POL(min(x_1)) = 0
 ||    POL(nil) = 0
 ||    POL(replace(x_1, x_2, x_3)) = x_3
 ||    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:
 || 
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 || 
 || 
 || ----------------------------------------
 || 
 || (43)
 || Obligation:
 || Q DP problem:
 || P is empty.
 || The TRS R consists of the following rules:
 || 
 ||    min(cons(0, nil)) -> 0
 ||    min(cons(s(%X), nil)) -> s(%X)
 ||    min(cons(%X, cons(%Y, %Z))) -> if!fac6220min(le(%X, %Y), cons(%X, cons(%Y, %Z)))
 ||    if!fac6220min(true, cons(%X, cons(%Y, %Z))) -> min(cons(%X, %Z))
 ||    if!fac6220min(false, cons(%X, cons(%Y, %Z))) -> min(cons(%Y, %Z))
 ||    replace(%X, %Y, nil) -> nil
 ||    replace(%X, %Y, cons(%Z, %U)) -> if!fac6220replace(eq(%X, %Z), %X, %Y, cons(%Z, %U))
 ||    eq(0, 0) -> true
 ||    eq(0, s(%X)) -> false
 ||    eq(s(%X), 0) -> false
 ||    eq(s(%X), s(%Y)) -> eq(%X, %Y)
 ||    if!fac6220replace(true, %X, %Y, cons(%Z, %U)) -> cons(%Y, %U)
 ||    if!fac6220replace(false, %X, %Y, cons(%Z, %U)) -> cons(%Z, replace(%X, %Y, %U))
 ||    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(cons(0, nil))
 ||    min(cons(s(x0), nil))
 ||    min(cons(x0, cons(x1, x2)))
 ||    if!fac6220min(true, cons(x0, cons(x1, x2)))
 ||    if!fac6220min(false, cons(x0, cons(x1, x2)))
 ||    replace(x0, x1, nil)
 ||    replace(x0, x1, cons(x2, x3))
 ||    if!fac6220replace(true, x0, x1, cons(x2, x3))
 ||    if!fac6220replace(false, x0, x1, cons(x2, x3))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (44) PisEmptyProof (EQUIVALENT)
 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
 || ----------------------------------------
 || 
 || (45)
 || 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, cons(X, Y)) =#> F(X)   
    1] map#(F, cons(X, Y)) =#> map#(F, Y)   
    2] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    3] filter#(F, cons(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) 
    min(cons(0, nil)) => 0 
    min(cons(s(X), nil)) => s(X) 
    min(cons(X, cons(Y, Z))) => if!fac6220min(le(X, Y), cons(X, cons(Y, Z))) 
    if!fac6220min(true, cons(X, cons(Y, Z))) => min(cons(X, Z)) 
    if!fac6220min(false, cons(X, cons(Y, Z))) => min(cons(Y, Z)) 
    replace(X, Y, nil) => nil 
    replace(X, Y, cons(Z, U)) => if!fac6220replace(eq(X, Z), X, Y, cons(Z, U)) 
    if!fac6220replace(true, X, Y, cons(Z, U)) => cons(Y, U) 
    if!fac6220replace(false, X, Y, cons(Z, U)) => cons(Z, replace(X, Y, U)) 
    sort(nil) => nil 
    sort(cons(X, Y)) => cons(min(cons(X, Y)), sort(replace(min(cons(X, Y)), X, Y))) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
    filter2(true, F, X, Y) => cons(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, cons(X, Y)) >? F(X) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) 
  filter#(F, cons(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) 
  min(cons(0, nil)) >= 0 
  min(cons(s(X), nil)) >= s(X) 
  min(cons(X, cons(Y, Z))) >= if!fac6220min(le(X, Y), cons(X, cons(Y, Z))) 
  if!fac6220min(true, cons(X, cons(Y, Z))) >= min(cons(X, Z)) 
  if!fac6220min(false, cons(X, cons(Y, Z))) >= min(cons(Y, Z)) 
  replace(X, Y, nil) >= nil 
  replace(X, Y, cons(Z, U)) >= if!fac6220replace(eq(X, Z), X, Y, cons(Z, U)) 
  if!fac6220replace(true, X, Y, cons(Z, U)) >= cons(Y, U) 
  if!fac6220replace(false, X, Y, cons(Z, U)) >= cons(Z, replace(X, Y, U)) 
  sort(nil) >= nil 
  sort(cons(X, Y)) >= cons(min(cons(X, Y)), sort(replace(min(cons(X, Y)), X, Y))) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  filter(F, nil) >= nil 
  filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) >= cons(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 = 0 
  cons = \y0y1.1 + y1 + 2y0 
  eq = \y0y1.2 
  false = 0 
  filter = \G0y1.2y1 
  filter2 = \y0G1y2y3.1 + 2y2 + 2y3 
  filter2# = \y0G1y2y3.G1(y3) 
  filter# = \G0y1.G0(y1) 
  if!fac6220min = \y0y1.0 
  if!fac6220replace = \y0y1y2y3.y3 + 2y2 
  le = \y0y1.0 
  map = \G0y1.2y1 + 2y1G0(y1) 
  map# = \G0y1.3 + 2G0(y1) 
  min = \y0.0 
  nil = 0 
  replace = \y0y1y2.y2 + 2y1 
  s = \y0.0 
  sort = \y0.y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 3 + 2F0(1 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 3 + 2F0(1 + x2 + 2x1) >= 3 + 2F0(x2) = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = F0(1 + x2 + 2x1) >= F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = F0(1 + x2 + 2x1) >= F0(x1) = [[_F0(_x1)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = F0(x2) >= F0(x2) = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = F0(x2) >= F0(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)]] = 0 >= 0 = [[true]] 
  [[le(s(_x0), 0)]] = 0 >= 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 0 >= 0 = [[le(_x0, _x1)]] 
  [[min(cons(0, nil))]] = 0 >= 0 = [[0]] 
  [[min(cons(s(_x0), nil))]] = 0 >= 0 = [[s(_x0)]] 
  [[min(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[if!fac6220min(le(_x0, _x1), cons(_x0, cons(_x1, _x2)))]] 
  [[if!fac6220min(true, cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[min(cons(_x0, _x2))]] 
  [[if!fac6220min(false, cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[min(cons(_x1, _x2))]] 
  [[replace(_x0, _x1, nil)]] = 2x1 >= 0 = [[nil]] 
  [[replace(_x0, _x1, cons(_x2, _x3))]] = 1 + x3 + 2x1 + 2x2 >= 1 + x3 + 2x1 + 2x2 = [[if!fac6220replace(eq(_x0, _x2), _x0, _x1, cons(_x2, _x3))]] 
  [[if!fac6220replace(true, _x0, _x1, cons(_x2, _x3))]] = 1 + x3 + 2x1 + 2x2 >= 1 + x3 + 2x1 = [[cons(_x1, _x3)]] 
  [[if!fac6220replace(false, _x0, _x1, cons(_x2, _x3))]] = 1 + x3 + 2x1 + 2x2 >= 1 + x3 + 2x1 + 2x2 = [[cons(_x2, replace(_x0, _x1, _x3))]] 
  [[sort(nil)]] = 0 >= 0 = [[nil]] 
  [[sort(cons(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[cons(min(cons(_x0, _x1)), sort(replace(min(cons(_x0, _x1)), _x0, _x1)))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + 2x2 + 4x1 + 2x2F0(1 + x2 + 2x1) + 2F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) >= 1 + 2x2 + 2x2F0(x2) + 2max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 + 2x2 + 4x1 >= 1 + 2x1 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 1 + 2x1 + 2x2 >= 1 + 2x1 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 1 + 2x1 + 2x2 >= 2x2 = [[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, cons(X, Y)) =#> map#(F, Y)   
  filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
  filter#(F, cons(X, Y)) =#> F(X)   
  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 : 3, 4 
    * 2 : 0, 1, 2, 3, 4 
    * 3 : 1, 2 
    * 4 : 1, 2 

This graph has the following strongly connected components:

  P_2:

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

  P_3:

    filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    filter#(F, cons(X, Y)) =#> F(X)   
    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 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:

  filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) 
  filter#(F, cons(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) 
  min(cons(0, nil)) >= 0 
  min(cons(s(X), nil)) >= s(X) 
  min(cons(X, cons(Y, Z))) >= if!fac6220min(le(X, Y), cons(X, cons(Y, Z))) 
  if!fac6220min(true, cons(X, cons(Y, Z))) >= min(cons(X, Z)) 
  if!fac6220min(false, cons(X, cons(Y, Z))) >= min(cons(Y, Z)) 
  replace(X, Y, nil) >= nil 
  replace(X, Y, cons(Z, U)) >= if!fac6220replace(eq(X, Z), X, Y, cons(Z, U)) 
  if!fac6220replace(true, X, Y, cons(Z, U)) >= cons(Y, U) 
  if!fac6220replace(false, X, Y, cons(Z, U)) >= cons(Z, replace(X, Y, U)) 
  sort(nil) >= nil 
  sort(cons(X, Y)) >= cons(min(cons(X, Y)), sort(replace(min(cons(X, Y)), X, Y))) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  filter(F, nil) >= nil 
  filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) >= cons(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 = 0 
  cons = \y0y1.1 + y0 + y1 
  eq = \y0y1.2 
  false = 0 
  filter = \G0y1.2y1 
  filter2 = \y0G1y2y3.2 + y2 + 2y3 
  filter2# = \y0G1y2y3.y3 + 2y3G1(y3) 
  filter# = \G0y1.y1 + 2y1G0(y1) 
  if!fac6220min = \y0y1.0 
  if!fac6220replace = \y0y1y2y3.y2 + y3 
  le = \y0y1.2 
  map = \G0y1.2y1 + y1G0(y1) 
  min = \y0.0 
  nil = 0 
  replace = \y0y1y2.y1 + y2 
  s = \y0.0 
  sort = \y0.2 + 2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[filter#(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > x2 + 2x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = x2 + 2x2F0(x2) >= x2 + 2x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = x2 + 2x2F0(x2) >= x2 + 2x2F0(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)]] = 2 >= 0 = [[true]] 
  [[le(s(_x0), 0)]] = 2 >= 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 2 >= 2 = [[le(_x0, _x1)]] 
  [[min(cons(0, nil))]] = 0 >= 0 = [[0]] 
  [[min(cons(s(_x0), nil))]] = 0 >= 0 = [[s(_x0)]] 
  [[min(cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[if!fac6220min(le(_x0, _x1), cons(_x0, cons(_x1, _x2)))]] 
  [[if!fac6220min(true, cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[min(cons(_x0, _x2))]] 
  [[if!fac6220min(false, cons(_x0, cons(_x1, _x2)))]] = 0 >= 0 = [[min(cons(_x1, _x2))]] 
  [[replace(_x0, _x1, nil)]] = x1 >= 0 = [[nil]] 
  [[replace(_x0, _x1, cons(_x2, _x3))]] = 1 + x1 + x2 + x3 >= 1 + x1 + x2 + x3 = [[if!fac6220replace(eq(_x0, _x2), _x0, _x1, cons(_x2, _x3))]] 
  [[if!fac6220replace(true, _x0, _x1, cons(_x2, _x3))]] = 1 + x1 + x2 + x3 >= 1 + x1 + x3 = [[cons(_x1, _x3)]] 
  [[if!fac6220replace(false, _x0, _x1, cons(_x2, _x3))]] = 1 + x1 + x2 + x3 >= 1 + x1 + x2 + x3 = [[cons(_x2, replace(_x0, _x1, _x3))]] 
  [[sort(nil)]] = 2 >= 0 = [[nil]] 
  [[sort(cons(_x0, _x1))]] = 4 + 2x0 + 2x1 >= 3 + 2x0 + 2x1 = [[cons(min(cons(_x0, _x1)), sort(replace(min(cons(_x0, _x1)), _x0, _x1)))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) >= 1 + 2x2 + x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 >= 2 + x1 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2 + x1 + 2x2 >= 1 + x1 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 + x1 + 2x2 >= 2x2 = [[filter(_F0, _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_0, minimal, formative) by (P_4, R_0, minimal, formative), where P_4 consists of:

  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, cons(X, Y))) = cons(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.