We consider the system Applicative_first_order_05__#3.57.

  Alphabet:

    0 : [] --> b 
    app : [c * c] --> c 
    cons : [b * c] --> c 
    false : [] --> a 
    filter : [b -> a * c] --> c 
    filter2 : [a * b -> a * b * c] --> c 
    map : [b -> b * c] --> c 
    minus : [b * b] --> b 
    nil : [] --> c 
    plus : [b * b] --> b 
    quot : [b * b] --> b 
    s : [b] --> b 
    sum : [c] --> c 
    true : [] --> a 

  Rules:

    minus(x, 0) => x 
    minus(s(x), s(y)) => minus(x, y) 
    minus(minus(x, y), z) => minus(x, plus(y, z)) 
    quot(0, s(x)) => 0 
    quot(s(x), s(y)) => s(quot(minus(x, y), s(y))) 
    plus(0, x) => x 
    plus(s(x), y) => s(plus(x, y)) 
    app(nil, x) => x 
    app(x, nil) => x 
    app(cons(x, y), z) => cons(x, app(y, z)) 
    sum(cons(x, nil)) => cons(x, nil) 
    sum(cons(x, cons(y, z))) => sum(cons(plus(x, y), z)) 
    sum(app(x, cons(y, cons(z, u)))) => sum(app(x, sum(cons(y, cons(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) 

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(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  minus(minus(X, Y), Z) => minus(X, plus(Y, Z)) 
  quot(0, s(X)) => 0 
  quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  app(nil, X) => X 
  app(X, nil) => X 
  app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
  sum(cons(X, nil)) => cons(X, nil) 
  sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) 
  sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(Z, U))))) 

Moreover, the system is finitely branching.  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 Ce-terminating when seen as a many-sorted first-order TRS.

According to the external first-order termination prover, this system is indeed Ce-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) DependencyPairsProof [EQUIVALENT]
 || (2) QDP
 || (3) DependencyGraphProof [EQUIVALENT]
 || (4) AND
 ||     (5) QDP
 ||         (6) UsableRulesProof [EQUIVALENT]
 ||         (7) QDP
 ||         (8) QDPSizeChangeProof [EQUIVALENT]
 ||         (9) YES
 ||     (10) QDP
 ||         (11) UsableRulesProof [EQUIVALENT]
 ||         (12) QDP
 ||         (13) QDPSizeChangeProof [EQUIVALENT]
 ||         (14) YES
 ||     (15) QDP
 ||         (16) UsableRulesProof [EQUIVALENT]
 ||         (17) QDP
 ||         (18) MNOCProof [EQUIVALENT]
 ||         (19) QDP
 ||         (20) UsableRulesReductionPairsProof [EQUIVALENT]
 ||         (21) QDP
 ||         (22) PisEmptyProof [EQUIVALENT]
 ||         (23) YES
 ||     (24) QDP
 ||         (25) UsableRulesProof [EQUIVALENT]
 ||         (26) QDP
 ||         (27) UsableRulesReductionPairsProof [EQUIVALENT]
 ||         (28) QDP
 ||         (29) MRRProof [EQUIVALENT]
 ||         (30) QDP
 ||         (31) DependencyGraphProof [EQUIVALENT]
 ||         (32) TRUE
 ||     (33) QDP
 ||         (34) UsableRulesProof [EQUIVALENT]
 ||         (35) QDP
 ||         (36) MNOCProof [EQUIVALENT]
 ||         (37) QDP
 ||         (38) QDPSizeChangeProof [EQUIVALENT]
 ||         (39) YES
 ||     (40) QDP
 ||         (41) QDPOrderProof [EQUIVALENT]
 ||         (42) QDP
 ||         (43) PisEmptyProof [EQUIVALENT]
 ||         (44) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(minus(%X, %Y), %Z) -> minus(%X, plus(%Y, %Z))
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    app(nil, %X) -> %X
 ||    app(%X, nil) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    sum(cons(%X, nil)) -> cons(%X, nil)
 ||    sum(cons(%X, cons(%Y, %Z))) -> sum(cons(plus(%X, %Y), %Z))
 ||    sum(app(%X, cons(%Y, cons(%Z, %U)))) -> sum(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) DependencyPairsProof (EQUIVALENT)
 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 ||    MINUS(minus(%X, %Y), %Z) -> MINUS(%X, plus(%Y, %Z))
 ||    MINUS(minus(%X, %Y), %Z) -> PLUS(%Y, %Z)
 ||    QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y))
 ||    QUOT(s(%X), s(%Y)) -> MINUS(%X, %Y)
 ||    PLUS(s(%X), %Y) -> PLUS(%X, %Y)
 ||    APP(cons(%X, %Y), %Z) -> APP(%Y, %Z)
 ||    SUM(cons(%X, cons(%Y, %Z))) -> SUM(cons(plus(%X, %Y), %Z))
 ||    SUM(cons(%X, cons(%Y, %Z))) -> PLUS(%X, %Y)
 ||    SUM(app(%X, cons(%Y, cons(%Z, %U)))) -> SUM(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 ||    SUM(app(%X, cons(%Y, cons(%Z, %U)))) -> APP(%X, sum(cons(%Y, cons(%Z, %U))))
 ||    SUM(app(%X, cons(%Y, cons(%Z, %U)))) -> SUM(cons(%Y, cons(%Z, %U)))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(minus(%X, %Y), %Z) -> minus(%X, plus(%Y, %Z))
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    app(nil, %X) -> %X
 ||    app(%X, nil) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    sum(cons(%X, nil)) -> cons(%X, nil)
 ||    sum(cons(%X, cons(%Y, %Z))) -> sum(cons(plus(%X, %Y), %Z))
 ||    sum(app(%X, cons(%Y, cons(%Z, %U)))) -> sum(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (3) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 5 less nodes.
 || ----------------------------------------
 || 
 || (4)
 || Complex Obligation (AND)
 || 
 || ----------------------------------------
 || 
 || (5)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    APP(cons(%X, %Y), %Z) -> APP(%Y, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(minus(%X, %Y), %Z) -> minus(%X, plus(%Y, %Z))
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    app(nil, %X) -> %X
 ||    app(%X, nil) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    sum(cons(%X, nil)) -> cons(%X, nil)
 ||    sum(cons(%X, cons(%Y, %Z))) -> sum(cons(plus(%X, %Y), %Z))
 ||    sum(app(%X, cons(%Y, cons(%Z, %U)))) -> sum(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (6) UsableRulesProof (EQUIVALENT)
 || We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (7)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    APP(cons(%X, %Y), %Z) -> APP(%Y, %Z)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (8) 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(cons(%X, %Y), %Z) -> APP(%Y, %Z)
 || The graph contains the following edges 1 > 1, 2 >= 2
 || 
 || 
 || ----------------------------------------
 || 
 || (9)
 || YES
 || 
 || ----------------------------------------
 || 
 || (10)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    PLUS(s(%X), %Y) -> PLUS(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(minus(%X, %Y), %Z) -> minus(%X, plus(%Y, %Z))
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    app(nil, %X) -> %X
 ||    app(%X, nil) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    sum(cons(%X, nil)) -> cons(%X, nil)
 ||    sum(cons(%X, cons(%Y, %Z))) -> sum(cons(plus(%X, %Y), %Z))
 ||    sum(app(%X, cons(%Y, cons(%Z, %U)))) -> sum(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (11) UsableRulesProof (EQUIVALENT)
 || We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (12)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    PLUS(s(%X), %Y) -> PLUS(%X, %Y)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (13) 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:
 || *PLUS(s(%X), %Y) -> PLUS(%X, %Y)
 || The graph contains the following edges 1 > 1, 2 >= 2
 || 
 || 
 || ----------------------------------------
 || 
 || (14)
 || YES
 || 
 || ----------------------------------------
 || 
 || (15)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    SUM(cons(%X, cons(%Y, %Z))) -> SUM(cons(plus(%X, %Y), %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(minus(%X, %Y), %Z) -> minus(%X, plus(%Y, %Z))
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    app(nil, %X) -> %X
 ||    app(%X, nil) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    sum(cons(%X, nil)) -> cons(%X, nil)
 ||    sum(cons(%X, cons(%Y, %Z))) -> sum(cons(plus(%X, %Y), %Z))
 ||    sum(app(%X, cons(%Y, cons(%Z, %U)))) -> sum(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (16) UsableRulesProof (EQUIVALENT)
 || We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (17)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    SUM(cons(%X, cons(%Y, %Z))) -> SUM(cons(plus(%X, %Y), %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (18) MNOCProof (EQUIVALENT)
 || We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
 || ----------------------------------------
 || 
 || (19)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    SUM(cons(%X, cons(%Y, %Z))) -> SUM(cons(plus(%X, %Y), %Z))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 || 
 || The set Q consists of the following terms:
 || 
 ||    plus(0, x0)
 ||    plus(s(x0), x1)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (20) UsableRulesReductionPairsProof (EQUIVALENT)
 || By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
 || 
 || The following dependency pairs can be deleted:
 || 
 ||    SUM(cons(%X, cons(%Y, %Z))) -> SUM(cons(plus(%X, %Y), %Z))
 || The following rules are removed from R:
 || 
 ||    plus(0, %X) -> %X
 || Used ordering: POLO with Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 0
 ||    POL(SUM(x_1)) = 2*x_1
 ||    POL(cons(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
 ||    POL(plus(x_1, x_2)) = x_1 + x_2
 ||    POL(s(x_1)) = x_1
 || 
 || 
 || ----------------------------------------
 || 
 || (21)
 || Obligation:
 || Q DP problem:
 || P is empty.
 || The TRS R consists of the following rules:
 || 
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 || 
 || The set Q consists of the following terms:
 || 
 ||    plus(0, x0)
 ||    plus(s(x0), x1)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (22) PisEmptyProof (EQUIVALENT)
 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
 || ----------------------------------------
 || 
 || (23)
 || YES
 || 
 || ----------------------------------------
 || 
 || (24)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    SUM(app(%X, cons(%Y, cons(%Z, %U)))) -> SUM(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(minus(%X, %Y), %Z) -> minus(%X, plus(%Y, %Z))
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    app(nil, %X) -> %X
 ||    app(%X, nil) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    sum(cons(%X, nil)) -> cons(%X, nil)
 ||    sum(cons(%X, cons(%Y, %Z))) -> sum(cons(plus(%X, %Y), %Z))
 ||    sum(app(%X, cons(%Y, cons(%Z, %U)))) -> sum(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (25) UsableRulesProof (EQUIVALENT)
 || We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (26)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    SUM(app(%X, cons(%Y, cons(%Z, %U)))) -> SUM(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    sum(cons(%X, cons(%Y, %Z))) -> sum(cons(plus(%X, %Y), %Z))
 ||    app(nil, %X) -> %X
 ||    app(%X, nil) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    sum(cons(%X, nil)) -> cons(%X, nil)
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (27) UsableRulesReductionPairsProof (EQUIVALENT)
 || By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
 || 
 || No dependency pairs are removed.
 || 
 || The following rules are removed from R:
 || 
 ||    app(nil, %X) -> %X
 ||    app(%X, nil) -> %X
 ||    plus(0, %X) -> %X
 || Used ordering: POLO with Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 0
 ||    POL(SUM(x_1)) = x_1
 ||    POL(app(x_1, x_2)) = 2 + 2*x_1 + x_2
 ||    POL(cons(x_1, x_2)) = 2*x_1 + x_2
 ||    POL(nil) = 0
 ||    POL(plus(x_1, x_2)) = x_1 + x_2
 ||    POL(s(x_1)) = x_1
 ||    POL(sum(x_1)) = x_1
 || 
 || 
 || ----------------------------------------
 || 
 || (28)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    SUM(app(%X, cons(%Y, cons(%Z, %U)))) -> SUM(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    sum(cons(%X, cons(%Y, %Z))) -> sum(cons(plus(%X, %Y), %Z))
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    sum(cons(%X, nil)) -> cons(%X, nil)
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (29) MRRProof (EQUIVALENT)
 || By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
 || 
 || 
 || Strictly oriented rules of the TRS R:
 || 
 ||    sum(cons(%X, cons(%Y, %Z))) -> sum(cons(plus(%X, %Y), %Z))
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 || 
 || Used ordering: Polynomial interpretation [POLO]:
 || 
 ||    POL(SUM(x_1)) = x_1
 ||    POL(app(x_1, x_2)) = 1 + 2*x_1 + 2*x_2
 ||    POL(cons(x_1, x_2)) = 1 + x_1 + x_2
 ||    POL(nil) = 0
 ||    POL(plus(x_1, x_2)) = x_1 + x_2
 ||    POL(s(x_1)) = x_1
 ||    POL(sum(x_1)) = x_1
 || 
 || 
 || ----------------------------------------
 || 
 || (30)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    SUM(app(%X, cons(%Y, cons(%Z, %U)))) -> SUM(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    sum(cons(%X, nil)) -> cons(%X, nil)
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (31) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
 || ----------------------------------------
 || 
 || (32)
 || TRUE
 || 
 || ----------------------------------------
 || 
 || (33)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(minus(%X, %Y), %Z) -> MINUS(%X, plus(%Y, %Z))
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(minus(%X, %Y), %Z) -> minus(%X, plus(%Y, %Z))
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    app(nil, %X) -> %X
 ||    app(%X, nil) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    sum(cons(%X, nil)) -> cons(%X, nil)
 ||    sum(cons(%X, cons(%Y, %Z))) -> sum(cons(plus(%X, %Y), %Z))
 ||    sum(app(%X, cons(%Y, cons(%Z, %U)))) -> sum(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (34) UsableRulesProof (EQUIVALENT)
 || We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (35)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(minus(%X, %Y), %Z) -> MINUS(%X, plus(%Y, %Z))
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (36) MNOCProof (EQUIVALENT)
 || We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
 || ----------------------------------------
 || 
 || (37)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(minus(%X, %Y), %Z) -> MINUS(%X, plus(%Y, %Z))
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 || 
 || The set Q consists of the following terms:
 || 
 ||    plus(0, x0)
 ||    plus(s(x0), x1)
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (38) 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(minus(%X, %Y), %Z) -> MINUS(%X, plus(%Y, %Z))
 || The graph contains the following edges 1 > 1
 || 
 || 
 || *MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || The graph contains the following edges 1 > 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (39)
 || YES
 || 
 || ----------------------------------------
 || 
 || (40)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(minus(%X, %Y), %Z) -> minus(%X, plus(%Y, %Z))
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    app(nil, %X) -> %X
 ||    app(%X, nil) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    sum(cons(%X, nil)) -> cons(%X, nil)
 ||    sum(cons(%X, cons(%Y, %Z))) -> sum(cons(plus(%X, %Y), %Z))
 ||    sum(app(%X, cons(%Y, cons(%Z, %U)))) -> sum(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (41) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y))
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Combined order from the following AFS and order.
 || QUOT(x1, x2)  =  QUOT(x1)
 || 
 || s(x1)  =  s(x1)
 || 
 || minus(x1, x2)  =  x1
 || 
 || 0  =  0
 || 
 || plus(x1, x2)  =  x2
 || 
 || 
 || Recursive path order with status [RPO].
 || Quasi-Precedence: QUOT_1 > s_1
 || 0 > s_1
 || 
 || Status: QUOT_1: [1]
 || s_1: multiset status
 || 0: multiset status
 || 
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(minus(%X, %Y), %Z) -> minus(%X, plus(%Y, %Z))
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 || 
 || 
 || ----------------------------------------
 || 
 || (42)
 || Obligation:
 || Q DP problem:
 || P is empty.
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    minus(minus(%X, %Y), %Z) -> minus(%X, plus(%Y, %Z))
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    plus(0, %X) -> %X
 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
 ||    app(nil, %X) -> %X
 ||    app(%X, nil) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    sum(cons(%X, nil)) -> cons(%X, nil)
 ||    sum(cons(%X, cons(%Y, %Z))) -> sum(cons(plus(%X, %Y), %Z))
 ||    sum(app(%X, cons(%Y, cons(%Z, %U)))) -> sum(app(%X, sum(cons(%Y, cons(%Z, %U)))))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (43) PisEmptyProof (EQUIVALENT)
 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
 || ----------------------------------------
 || 
 || (44)
 || 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(X, 0) => X 
    minus(s(X), s(Y)) => minus(X, Y) 
    minus(minus(X, Y), Z) => minus(X, plus(Y, Z)) 
    quot(0, s(X)) => 0 
    quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    app(nil, X) => X 
    app(X, nil) => X 
    app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
    sum(cons(X, nil)) => cons(X, nil) 
    sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) 
    sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(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 ::=

  app(nil, X) => X 
  app(X, nil) => X 
  app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
  sum(cons(X, nil)) => cons(X, nil) 
  sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) 
  sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(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) 

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) 
  app(nil, X) >= X 
  app(X, nil) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  sum(cons(X, nil)) >= cons(X, nil) 
  sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) 
  sum(app(X, cons(Y, cons(Z, U)))) >= sum(app(X, sum(cons(Y, cons(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) 

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

The following interpretation satisfies the requirements:

  app = \y0y1.y1 + 3y0 
  cons = \y0y1.2 + y0 + y1 
  false = 3 
  filter = \G0y1.y1 
  filter2 = \y0G1y2y3.2 + y2 + y3 
  filter2# = \y0G1y2y3.2y3 + G1(y3) 
  filter# = \G0y1.2y1 + G0(y1) 
  map = \G0y1.y1 + y1G0(y1) 
  map# = \G0y1.3 + y1 + G0(y1) 
  nil = 0 
  plus = \y0y1.0 
  sum = \y0.y0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 5 + x1 + x2 + F0(2 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 5 + x1 + x2 + F0(2 + x1 + x2) > 3 + x2 + F0(x2) = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(2 + x1 + x2) > 2x2 + F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(2 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = 2x2 + F0(x2) >= 2x2 + F0(x2) = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = 2x2 + F0(x2) >= 2x2 + F0(x2) = [[filter#(_F0, _x2)]] 
  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(_x0, nil)]] = 3x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 6 + x2 + 3x0 + 3x1 >= 2 + x0 + x2 + 3x1 = [[cons(_x0, app(_x1, _x2))]] 
  [[sum(cons(_x0, nil))]] = 2 + x0 >= 2 + x0 = [[cons(_x0, nil)]] 
  [[sum(cons(_x0, cons(_x1, _x2)))]] = 4 + x0 + x1 + x2 >= 2 + x2 = [[sum(cons(plus(_x0, _x1), _x2))]] 
  [[sum(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 4 + x1 + x2 + x3 + 3x0 >= 4 + x1 + x2 + x3 + 3x0 = [[sum(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) >= 2 + x2 + x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 2 + 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_0, R_1, minimal, formative) by (P_1, R_1, minimal, formative), where P_1 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 (P_1, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_1, 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 +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.