We consider the system Applicative_first_order_05__perfect2.

  Alphabet:

    0 : [] --> a 
    cons : [c * d] --> d 
    f : [a * a * a * a] --> b 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    if : [b * b * b] --> b 
    le : [a * a] --> b 
    map : [c -> c * d] --> d 
    minus : [a * a] --> a 
    nil : [] --> d 
    perfectp : [a] --> b 
    s : [a] --> a 
    true : [] --> b 

  Rules:

    minus(0, x) => 0 
    minus(s(x), 0) => s(x) 
    minus(s(x), s(y)) => minus(x, y) 
    le(0, x) => true 
    le(s(x), 0) => false 
    le(s(x), s(y)) => le(x, y) 
    if(true, x, y) => x 
    if(false, x, y) => y 
    perfectp(0) => false 
    perfectp(s(x)) => f(x, s(0), s(x), s(x)) 
    f(0, x, 0, y) => true 
    f(0, x, s(y), z) => false 
    f(s(x), 0, y, z) => f(x, z, minus(y, s(x)), z) 
    f(s(x), s(y), z, u) => if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) 
    map(g, nil) => nil 
    map(g, cons(x, y)) => cons(g x, map(g, y)) 
    filter(g, nil) => nil 
    filter(g, cons(x, y)) => filter2(g x, g, x, y) 
    filter2(true, g, x, y) => cons(x, filter(g, y)) 
    filter2(false, g, x, y) => filter(g, 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:

  minus(0, X) => 0 
  minus(s(X), 0) => s(X) 
  minus(s(X), s(Y)) => minus(X, Y) 
  le(0, X) => true 
  le(s(X), 0) => false 
  le(s(X), s(Y)) => le(X, Y) 
  if(true, X, Y) => X 
  if(false, X, Y) => Y 
  perfectp(0) => false 
  perfectp(s(X)) => f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) => true 
  f(0, X, s(Y), Z) => false 
  f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) 
  f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 

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) QDPOrderProof [EQUIVALENT]
 ||         (29) QDP
 ||         (30) PisEmptyProof [EQUIVALENT]
 ||         (31) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    minus(0, %X) -> 0
 ||    minus(s(%X), 0) -> s(%X)
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    perfectp(0) -> false
 ||    perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X))
 ||    f(0, %X, 0, %Y) -> true
 ||    f(0, %X, s(%Y), %Z) -> false
 ||    f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U))
 || 
 || 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:
 || 
 ||    minus(0, %X) -> 0
 ||    minus(s(%X), 0) -> s(%X)
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    perfectp(0) -> false
 ||    perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X))
 ||    f(0, %X, 0, %Y) -> true
 ||    f(0, %X, s(%Y), %Z) -> false
 ||    f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(0, x0)
 ||    minus(s(x0), 0)
 ||    minus(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || 
 || ----------------------------------------
 || 
 || (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:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 ||    PERFECTP(s(%X)) -> F(%X, s(0), s(%X), s(%X))
 ||    F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    F(s(%X), 0, %Y, %Z) -> MINUS(%Y, s(%X))
 ||    F(s(%X), s(%Y), %Z, %U) -> IF(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U))
 ||    F(s(%X), s(%Y), %Z, %U) -> LE(%X, %Y)
 ||    F(s(%X), s(%Y), %Z, %U) -> F(s(%X), minus(%Y, %X), %Z, %U)
 ||    F(s(%X), s(%Y), %Z, %U) -> MINUS(%Y, %X)
 ||    F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(0, %X) -> 0
 ||    minus(s(%X), 0) -> s(%X)
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    perfectp(0) -> false
 ||    perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X))
 ||    f(0, %X, 0, %Y) -> true
 ||    f(0, %X, s(%Y), %Z) -> false
 ||    f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(0, x0)
 ||    minus(s(x0), 0)
 ||    minus(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (5) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 5 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:
 || 
 ||    minus(0, %X) -> 0
 ||    minus(s(%X), 0) -> s(%X)
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    perfectp(0) -> false
 ||    perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X))
 ||    f(0, %X, 0, %Y) -> true
 ||    f(0, %X, s(%Y), %Z) -> false
 ||    f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(0, x0)
 ||    minus(s(x0), 0)
 ||    minus(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || 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:
 || 
 ||    minus(0, x0)
 ||    minus(s(x0), 0)
 ||    minus(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || 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].
 || 
 ||    minus(0, x0)
 ||    minus(s(x0), 0)
 ||    minus(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || 
 || ----------------------------------------
 || 
 || (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:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(0, %X) -> 0
 ||    minus(s(%X), 0) -> s(%X)
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    perfectp(0) -> false
 ||    perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X))
 ||    f(0, %X, 0, %Y) -> true
 ||    f(0, %X, s(%Y), %Z) -> false
 ||    f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(0, x0)
 ||    minus(s(x0), 0)
 ||    minus(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || 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:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    minus(0, x0)
 ||    minus(s(x0), 0)
 ||    minus(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || 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].
 || 
 ||    minus(0, x0)
 ||    minus(s(x0), 0)
 ||    minus(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || 
 || ----------------------------------------
 || 
 || (18)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%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:
 || *MINUS(s(%X), s(%Y)) -> MINUS(%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:
 || 
 ||    F(s(%X), s(%Y), %Z, %U) -> F(s(%X), minus(%Y, %X), %Z, %U)
 ||    F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(0, %X) -> 0
 ||    minus(s(%X), 0) -> s(%X)
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    perfectp(0) -> false
 ||    perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X))
 ||    f(0, %X, 0, %Y) -> true
 ||    f(0, %X, s(%Y), %Z) -> false
 ||    f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(0, x0)
 ||    minus(s(x0), 0)
 ||    minus(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || 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:
 || 
 ||    F(s(%X), s(%Y), %Z, %U) -> F(s(%X), minus(%Y, %X), %Z, %U)
 ||    F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(0, %X) -> 0
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(s(%X), 0) -> s(%X)
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(0, x0)
 ||    minus(s(x0), 0)
 ||    minus(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || 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].
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    perfectp(0)
 ||    perfectp(s(x0))
 ||    f(0, x0, 0, x1)
 ||    f(0, x0, s(x1), x2)
 ||    f(s(x0), 0, x1, x2)
 ||    f(s(x0), s(x1), x2, x3)
 || 
 || 
 || ----------------------------------------
 || 
 || (25)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    F(s(%X), s(%Y), %Z, %U) -> F(s(%X), minus(%Y, %X), %Z, %U)
 ||    F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(0, %X) -> 0
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(s(%X), 0) -> s(%X)
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(0, x0)
 ||    minus(s(x0), 0)
 ||    minus(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.
 || 
 ||    F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z)
 ||    F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U)
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 0
 ||    POL(F(x_1, x_2, x_3, x_4)) = x_1
 ||    POL(minus(x_1, x_2)) = 0
 ||    POL(s(x_1)) = 1 + x_1
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || none
 || 
 || 
 || ----------------------------------------
 || 
 || (27)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    F(s(%X), s(%Y), %Z, %U) -> F(s(%X), minus(%Y, %X), %Z, %U)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(0, %X) -> 0
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(s(%X), 0) -> s(%X)
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(0, x0)
 ||    minus(s(x0), 0)
 ||    minus(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (28) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    F(s(%X), s(%Y), %Z, %U) -> F(s(%X), minus(%Y, %X), %Z, %U)
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
 || 
 || POL( F_4(x_1, ..., x_4) ) = max{0, 2x_2 + x_3 + x_4 - 1}
 || POL( minus_2(x_1, x_2) ) = x_1
 || POL( 0 ) = 1
 || POL( s_1(x_1) ) = x_1 + 1
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    minus(0, %X) -> 0
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(s(%X), 0) -> s(%X)
 || 
 || 
 || ----------------------------------------
 || 
 || (29)
 || Obligation:
 || Q DP problem:
 || P is empty.
 || The TRS R consists of the following rules:
 || 
 ||    minus(0, %X) -> 0
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(s(%X), 0) -> s(%X)
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(0, x0)
 ||    minus(s(x0), 0)
 ||    minus(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (30) PisEmptyProof (EQUIVALENT)
 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
 || ----------------------------------------
 || 
 || (31)
 || 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:

    minus(0, X) => 0 
    minus(s(X), 0) => s(X) 
    minus(s(X), s(Y)) => minus(X, Y) 
    le(0, X) => true 
    le(s(X), 0) => false 
    le(s(X), s(Y)) => le(X, Y) 
    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    perfectp(0) => false 
    perfectp(s(X)) => f(X, s(0), s(X), s(X)) 
    f(0, X, 0, Y) => true 
    f(0, X, s(Y), Z) => false 
    f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) 
    f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 
    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).

The formative rules of (P_0, R_0) are R_1 ::=

  minus(0, X) => 0 
  minus(s(X), 0) => s(X) 
  minus(s(X), s(Y)) => minus(X, Y) 
  le(0, X) => true 
  le(s(X), 0) => false 
  le(s(X), s(Y)) => le(X, Y) 
  if(true, X, Y) => X 
  if(false, X, Y) => Y 
  perfectp(0) => false 
  perfectp(s(X)) => f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) => true 
  f(0, X, s(Y), Z) => false 
  f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) 
  f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  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) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_0, R_1, minimal, formative).

Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_1, 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) 
  minus(0, X) >= 0 
  minus(s(X), 0) >= s(X) 
  minus(s(X), s(Y)) >= minus(X, Y) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  perfectp(0) >= false 
  perfectp(s(X)) >= f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) >= true 
  f(0, X, s(Y), Z) >= false 
  f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) 
  f(s(X), s(Y), Z, U) >= if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  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 = \y0y1.2 + y1 + 2y0 
  f = \y0y1y2y3.0 
  false = 0 
  filter = \G0y1.2y1 + G0(y1) + 2y1G0(y1) 
  filter2 = \y0G1y2y3.2 + 2y0 + 2y2 + 2y3 + G1(y3) + 2y3G1(y3) 
  filter2# = \y0G1y2y3.G1(0) + y3G1(y3) 
  filter# = \G0y1.G0(0) + y1G0(y1) 
  if = \y0y1y2.y1 + y2 
  le = \y0y1.3 
  map = \G0y1.2y1 + 2y1G0(y1) + 2G0(y1) 
  map# = \G0y1.3 + y1 + G0(y1) + y1G0(y1) 
  minus = \y0y1.3 + 2y0 + 2y1 
  perfectp = \y0.3 
  s = \y0.2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 5 + x2 + 2x1 + 2x1F0(2 + x2 + 2x1) + 3F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 5 + x2 + 2x1 + 2x1F0(2 + x2 + 2x1) + 3F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > 3 + x2 + F0(x2) + x2F0(x2) = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = F0(0) + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) >= F0(0) + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = F0(0) + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) >= F0(x1) = [[_F0(_x1)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = F0(0) + x2F0(x2) >= F0(0) + x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = F0(0) + x2F0(x2) >= F0(0) + x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[minus(0, _x0)]] = 3 + 2x0 >= 0 = [[0]] 
  [[minus(s(_x0), 0)]] = 3 + 4x0 >= 2x0 = [[s(_x0)]] 
  [[minus(s(_x0), s(_x1))]] = 3 + 4x0 + 4x1 >= 3 + 2x0 + 2x1 = [[minus(_x0, _x1)]] 
  [[le(0, _x0)]] = 3 >= 0 = [[true]] 
  [[le(s(_x0), 0)]] = 3 >= 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 3 >= 3 = [[le(_x0, _x1)]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[perfectp(0)]] = 3 >= 0 = [[false]] 
  [[perfectp(s(_x0))]] = 3 >= 0 = [[f(_x0, s(0), s(_x0), s(_x0))]] 
  [[f(0, _x0, 0, _x1)]] = 0 >= 0 = [[true]] 
  [[f(0, _x0, s(_x1), _x2)]] = 0 >= 0 = [[false]] 
  [[f(s(_x0), 0, _x1, _x2)]] = 0 >= 0 = [[f(_x0, _x2, minus(_x1, s(_x0)), _x2)]] 
  [[f(s(_x0), s(_x1), _x2, _x3)]] = 0 >= 0 = [[if(le(_x0, _x1), f(s(_x0), minus(_x1, _x0), _x2, _x3), f(_x0, _x3, _x2, _x3))]] 
  [[map(_F0, cons(_x1, _x2))]] = 4 + 2x2 + 4x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 6F0(2 + x2 + 2x1) >= 2 + 2x2 + 2x2F0(x2) + 2F0(x2) + 2max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x2 + 4x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 5F0(2 + x2 + 2x1) >= 2 + 2x1 + 2x2 + F0(x2) + 2x2F0(x2) + 2max(x1, F0(x1)) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2 + 2x1 + 2x2 + F0(x2) + 2x2F0(x2) >= 2 + 2x1 + 2x2 + F0(x2) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 + 2x1 + 2x2 + F0(x2) + 2x2F0(x2) >= 2x2 + 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_1, minimal, formative) by (P_1, R_1, minimal, formative), where P_1 consists of:

  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_1, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_1, 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) 
  minus(0, X) >= 0 
  minus(s(X), 0) >= s(X) 
  minus(s(X), s(Y)) >= minus(X, Y) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  perfectp(0) >= false 
  perfectp(s(X)) >= f(X, s(0), s(X), s(X)) 
  f(0, X, 0, Y) >= true 
  f(0, X, s(Y), Z) >= false 
  f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) 
  f(s(X), s(Y), Z, U) >= if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  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 = 1 
  cons = \y0y1.1 + y0 + y1 
  f = \y0y1y2y3.0 
  false = 0 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.1 + y2 + y3 
  filter2# = \y0G1y2y3.1 + y3G1(y3) 
  filter# = \G0y1.1 + y1G0(y1) 
  if = \y0y1y2.y1 + y2 
  le = \y0y1.3 
  map = \G0y1.y1 + 2y1G0(y1) 
  minus = \y0y1.3 + 2y0 + 2y1 
  perfectp = \y0.3 + y0 
  s = \y0.2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[filter#(_F0, cons(_x1, _x2))]] = 1 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) >= 1 + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 1 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) > 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)]] 
  [[minus(0, _x0)]] = 5 + 2x0 >= 1 = [[0]] 
  [[minus(s(_x0), 0)]] = 5 + 4x0 >= 2x0 = [[s(_x0)]] 
  [[minus(s(_x0), s(_x1))]] = 3 + 4x0 + 4x1 >= 3 + 2x0 + 2x1 = [[minus(_x0, _x1)]] 
  [[le(0, _x0)]] = 3 >= 0 = [[true]] 
  [[le(s(_x0), 0)]] = 3 >= 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 3 >= 3 = [[le(_x0, _x1)]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[perfectp(0)]] = 4 >= 0 = [[false]] 
  [[perfectp(s(_x0))]] = 3 + 2x0 >= 0 = [[f(_x0, s(0), s(_x0), s(_x0))]] 
  [[f(0, _x0, 0, _x1)]] = 0 >= 0 = [[true]] 
  [[f(0, _x0, s(_x1), _x2)]] = 0 >= 0 = [[false]] 
  [[f(s(_x0), 0, _x1, _x2)]] = 0 >= 0 = [[f(_x0, _x2, minus(_x1, s(_x0)), _x2)]] 
  [[f(s(_x0), s(_x1), _x2, _x3)]] = 0 >= 0 = [[if(le(_x0, _x1), f(s(_x0), minus(_x1, _x0), _x2, _x3), f(_x0, _x3, _x2, _x3))]] 
  [[map(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) >= 1 + x2 + 2x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 1 + x1 + x2 >= 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_1, R_1, minimal, formative) by (P_2, R_1, minimal, formative), where P_2 consists of:

  filter#(F, cons(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_2, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_1, 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, cons(X, Y))) = cons(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_2, R_1, minimal, f) by (P_3, R_1, minimal, f), where P_3 contains:

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

Thus, the original system is terminating if (P_3, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_1, 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.

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.