We consider the system Applicative_05__Ex7_9.

  Alphabet:

    0 : [] --> a 
    cons : [b * c] --> c 
    d : [a * a] --> c 
    false : [] --> c 
    filter : [b -> c * c] --> c 
    gtr : [a * a] --> c 
    if : [c * c * c] --> c 
    len : [c] --> a 
    nil : [] --> c 
    s : [a] --> a 
    sub : [a * a] --> a 
    true : [] --> c 

  Rules:

    if(true, x, y) => x 
    if(false, x, y) => y 
    sub(x, 0) => x 
    sub(s(x), s(y)) => sub(x, y) 
    gtr(0, x) => false 
    gtr(s(x), 0) => true 
    gtr(s(x), s(y)) => gtr(x, y) 
    d(x, 0) => true 
    d(s(x), s(y)) => if(gtr(x, y), false, d(s(x), sub(y, x))) 
    len(nil) => 0 
    len(cons(x, y)) => s(len(y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => if(f x, cons(x, filter(f, 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:

  if(true, X, Y) => X 
  if(false, X, Y) => Y 
  sub(X, 0) => X 
  sub(s(X), s(Y)) => sub(X, Y) 
  gtr(0, X) => false 
  gtr(s(X), 0) => true 
  gtr(s(X), s(Y)) => gtr(X, Y) 
  d(X, 0) => true 
  d(s(X), s(Y)) => if(gtr(X, Y), false, d(s(X), sub(Y, X))) 
  len(nil) => 0 
  len(cons(X, Y)) => s(len(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) QDPSizeChangeProof [EQUIVALENT]
 ||         (20) YES
 ||     (21) QDP
 ||         (22) UsableRulesProof [EQUIVALENT]
 ||         (23) QDP
 ||         (24) QReductionProof [EQUIVALENT]
 ||         (25) QDP
 ||         (26) QDPSizeChangeProof [EQUIVALENT]
 ||         (27) YES
 ||     (28) QDP
 ||         (29) UsableRulesProof [EQUIVALENT]
 ||         (30) QDP
 ||         (31) QReductionProof [EQUIVALENT]
 ||         (32) QDP
 ||         (33) QDPOrderProof [EQUIVALENT]
 ||         (34) QDP
 ||         (35) PisEmptyProof [EQUIVALENT]
 ||         (36) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    sub(%X, 0) -> %X
 ||    sub(s(%X), s(%Y)) -> sub(%X, %Y)
 ||    gtr(0, %X) -> false
 ||    gtr(s(%X), 0) -> true
 ||    gtr(s(%X), s(%Y)) -> gtr(%X, %Y)
 ||    d(%X, 0) -> true
 ||    d(s(%X), s(%Y)) -> if(gtr(%X, %Y), false, d(s(%X), sub(%Y, %X)))
 ||    len(nil) -> 0
 ||    len(cons(%X, %Y)) -> s(len(%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:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    sub(%X, 0) -> %X
 ||    sub(s(%X), s(%Y)) -> sub(%X, %Y)
 ||    gtr(0, %X) -> false
 ||    gtr(s(%X), 0) -> true
 ||    gtr(s(%X), s(%Y)) -> gtr(%X, %Y)
 ||    d(%X, 0) -> true
 ||    d(s(%X), s(%Y)) -> if(gtr(%X, %Y), false, d(s(%X), sub(%Y, %X)))
 ||    len(nil) -> 0
 ||    len(cons(%X, %Y)) -> s(len(%Y))
 || 
 || The set Q consists of the following terms:
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(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:
 || 
 ||    SUB(s(%X), s(%Y)) -> SUB(%X, %Y)
 ||    GTR(s(%X), s(%Y)) -> GTR(%X, %Y)
 ||    D(s(%X), s(%Y)) -> IF(gtr(%X, %Y), false, d(s(%X), sub(%Y, %X)))
 ||    D(s(%X), s(%Y)) -> GTR(%X, %Y)
 ||    D(s(%X), s(%Y)) -> D(s(%X), sub(%Y, %X))
 ||    D(s(%X), s(%Y)) -> SUB(%Y, %X)
 ||    LEN(cons(%X, %Y)) -> LEN(%Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    sub(%X, 0) -> %X
 ||    sub(s(%X), s(%Y)) -> sub(%X, %Y)
 ||    gtr(0, %X) -> false
 ||    gtr(s(%X), 0) -> true
 ||    gtr(s(%X), s(%Y)) -> gtr(%X, %Y)
 ||    d(%X, 0) -> true
 ||    d(s(%X), s(%Y)) -> if(gtr(%X, %Y), false, d(s(%X), sub(%Y, %X)))
 ||    len(nil) -> 0
 ||    len(cons(%X, %Y)) -> s(len(%Y))
 || 
 || The set Q consists of the following terms:
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(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 4 SCCs with 3 less nodes.
 || ----------------------------------------
 || 
 || (6)
 || Complex Obligation (AND)
 || 
 || ----------------------------------------
 || 
 || (7)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    LEN(cons(%X, %Y)) -> LEN(%Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    sub(%X, 0) -> %X
 ||    sub(s(%X), s(%Y)) -> sub(%X, %Y)
 ||    gtr(0, %X) -> false
 ||    gtr(s(%X), 0) -> true
 ||    gtr(s(%X), s(%Y)) -> gtr(%X, %Y)
 ||    d(%X, 0) -> true
 ||    d(s(%X), s(%Y)) -> if(gtr(%X, %Y), false, d(s(%X), sub(%Y, %X)))
 ||    len(nil) -> 0
 ||    len(cons(%X, %Y)) -> s(len(%Y))
 || 
 || The set Q consists of the following terms:
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(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:
 || 
 ||    LEN(cons(%X, %Y)) -> LEN(%Y)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(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].
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (11)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    LEN(cons(%X, %Y)) -> LEN(%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:
 || *LEN(cons(%X, %Y)) -> LEN(%Y)
 || The graph contains the following edges 1 > 1
 || 
 || 
 || ----------------------------------------
 || 
 || (13)
 || YES
 || 
 || ----------------------------------------
 || 
 || (14)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    GTR(s(%X), s(%Y)) -> GTR(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    sub(%X, 0) -> %X
 ||    sub(s(%X), s(%Y)) -> sub(%X, %Y)
 ||    gtr(0, %X) -> false
 ||    gtr(s(%X), 0) -> true
 ||    gtr(s(%X), s(%Y)) -> gtr(%X, %Y)
 ||    d(%X, 0) -> true
 ||    d(s(%X), s(%Y)) -> if(gtr(%X, %Y), false, d(s(%X), sub(%Y, %X)))
 ||    len(nil) -> 0
 ||    len(cons(%X, %Y)) -> s(len(%Y))
 || 
 || The set Q consists of the following terms:
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(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:
 || 
 ||    GTR(s(%X), s(%Y)) -> GTR(%X, %Y)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(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].
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (18)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    GTR(s(%X), s(%Y)) -> GTR(%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:
 || *GTR(s(%X), s(%Y)) -> GTR(%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:
 || 
 ||    SUB(s(%X), s(%Y)) -> SUB(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    sub(%X, 0) -> %X
 ||    sub(s(%X), s(%Y)) -> sub(%X, %Y)
 ||    gtr(0, %X) -> false
 ||    gtr(s(%X), 0) -> true
 ||    gtr(s(%X), s(%Y)) -> gtr(%X, %Y)
 ||    d(%X, 0) -> true
 ||    d(s(%X), s(%Y)) -> if(gtr(%X, %Y), false, d(s(%X), sub(%Y, %X)))
 ||    len(nil) -> 0
 ||    len(cons(%X, %Y)) -> s(len(%Y))
 || 
 || The set Q consists of the following terms:
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(cons(x0, x1))
 || 
 || 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:
 || 
 ||    SUB(s(%X), s(%Y)) -> SUB(%X, %Y)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(cons(x0, x1))
 || 
 || 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].
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (25)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    SUB(s(%X), s(%Y)) -> SUB(%X, %Y)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (26) 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:
 || *SUB(s(%X), s(%Y)) -> SUB(%X, %Y)
 || The graph contains the following edges 1 > 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (27)
 || YES
 || 
 || ----------------------------------------
 || 
 || (28)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    D(s(%X), s(%Y)) -> D(s(%X), sub(%Y, %X))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    sub(%X, 0) -> %X
 ||    sub(s(%X), s(%Y)) -> sub(%X, %Y)
 ||    gtr(0, %X) -> false
 ||    gtr(s(%X), 0) -> true
 ||    gtr(s(%X), s(%Y)) -> gtr(%X, %Y)
 ||    d(%X, 0) -> true
 ||    d(s(%X), s(%Y)) -> if(gtr(%X, %Y), false, d(s(%X), sub(%Y, %X)))
 ||    len(nil) -> 0
 ||    len(cons(%X, %Y)) -> s(len(%Y))
 || 
 || The set Q consists of the following terms:
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (29) 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.
 || ----------------------------------------
 || 
 || (30)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    D(s(%X), s(%Y)) -> D(s(%X), sub(%Y, %X))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    sub(%X, 0) -> %X
 ||    sub(s(%X), s(%Y)) -> sub(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(cons(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (31) 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].
 || 
 ||    if(true, x0, x1)
 ||    if(false, x0, x1)
 ||    gtr(0, x0)
 ||    gtr(s(x0), 0)
 ||    gtr(s(x0), s(x1))
 ||    d(x0, 0)
 ||    d(s(x0), s(x1))
 ||    len(nil)
 ||    len(cons(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (32)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    D(s(%X), s(%Y)) -> D(s(%X), sub(%Y, %X))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    sub(%X, 0) -> %X
 ||    sub(s(%X), s(%Y)) -> sub(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (33) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    D(s(%X), s(%Y)) -> D(s(%X), sub(%Y, %X))
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
 || 
 || POL( D_2(x_1, x_2) ) = 2x_1 + 2x_2 + 2
 || POL( sub_2(x_1, x_2) ) = 2x_1
 || POL( 0 ) = 1
 || POL( s_1(x_1) ) = 2x_1 + 1
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    sub(%X, 0) -> %X
 ||    sub(s(%X), s(%Y)) -> sub(%X, %Y)
 || 
 || 
 || ----------------------------------------
 || 
 || (34)
 || Obligation:
 || Q DP problem:
 || P is empty.
 || The TRS R consists of the following rules:
 || 
 ||    sub(%X, 0) -> %X
 ||    sub(s(%X), s(%Y)) -> sub(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    sub(x0, 0)
 ||    sub(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (35) PisEmptyProof (EQUIVALENT)
 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
 || ----------------------------------------
 || 
 || (36)
 || 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] filter#(F, cons(X, Y)) =#> if#(F X, cons(X, filter(F, Y)), filter(F, Y))   
    1] filter#(F, cons(X, Y)) =#> F(X)   
    2] filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    3] filter#(F, cons(X, Y)) =#> filter#(F, Y)   

  Rules R_0:

    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    sub(X, 0) => X 
    sub(s(X), s(Y)) => sub(X, Y) 
    gtr(0, X) => false 
    gtr(s(X), 0) => true 
    gtr(s(X), s(Y)) => gtr(X, Y) 
    d(X, 0) => true 
    d(s(X), s(Y)) => if(gtr(X, Y), false, d(s(X), sub(Y, X))) 
    len(nil) => 0 
    len(cons(X, Y)) => s(len(Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => if(F X, cons(X, filter(F, 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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

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

This graph has the following strongly connected components:

  P_1:

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

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f).

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 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)) >? F(X) 
  filter#(F, cons(X, Y)) >? filter#(F, Y) 
  filter#(F, cons(X, Y)) >? filter#(F, Y) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  sub(X, 0) >= X 
  sub(s(X), s(Y)) >= sub(X, Y) 
  gtr(0, X) >= false 
  gtr(s(X), 0) >= true 
  gtr(s(X), s(Y)) >= gtr(X, Y) 
  d(X, 0) >= true 
  d(s(X), s(Y)) >= if(gtr(X, Y), false, d(s(X), sub(Y, X))) 
  len(nil) >= 0 
  len(cons(X, Y)) >= s(len(Y)) 
  filter(F, nil) >= nil 
  filter(F, cons(X, Y)) >= if(F X, cons(X, filter(F, Y)), filter(F, Y)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[0]] = _|_ 
  [[false]] = _|_ 
  [[gtr(x_1, x_2)]] = gtr 
  [[if(x_1, x_2, x_3)]] = if(x_2, x_3) 
  [[nil]] = _|_ 
  [[sub(x_1, x_2)]] = sub(x_1) 
  [[true]] = _|_ 

We choose Lex = {} and Mul = {@_{o -> o}, cons, d, filter, filter#, gtr, if, len, s, sub}, and the following precedence: gtr > filter > cons > len = s > d > filter# > if > @_{o -> o} > sub

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  filter#(F, cons(X, Y)) >= @_{o -> o}(F, X) 
  filter#(F, cons(X, Y)) > filter#(F, Y) 
  filter#(F, cons(X, Y)) >= filter#(F, Y) 
  if(X, Y) >= X 
  if(X, Y) >= Y 
  sub(X) >= X 
  sub(s(X)) >= sub(X) 
  gtr >= _|_ 
  gtr >= _|_ 
  gtr >= gtr 
  d(X, _|_) >= _|_ 
  d(s(X), s(Y)) >= if(_|_, d(s(X), sub(Y))) 
  len(_|_) >= _|_ 
  len(cons(X, Y)) >= s(len(Y)) 
  filter(F, _|_) >= _|_ 
  filter(F, cons(X, Y)) >= if(cons(X, filter(F, Y)), filter(F, Y)) 

With these choices, we have:

  1] filter#(F, cons(X, Y)) >= @_{o -> o}(F, X)  because [2], by (Star) 
  2] filter#*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter# > @_{o -> o}, [3] and [5], by (Copy) 
  3] filter#*(F, cons(X, Y)) >= F  because [4], by (Select) 
  4] F >= F  by (Meta) 
  5] filter#*(F, cons(X, Y)) >= X  because [6], by (Select) 
  6] cons(X, Y) >= X  because [7], by (Star) 
  7] cons*(X, Y) >= X  because [8], by (Select) 
  8] X >= X  by (Meta) 

  9] filter#(F, cons(X, Y)) > filter#(F, Y)  because [10], by definition 
  10] filter#*(F, cons(X, Y)) >= filter#(F, Y)  because filter# in Mul, [11] and [12], by (Stat) 
  11] F >= F  by (Meta) 
  12] cons(X, Y) > Y  because [13], by definition 
  13] cons*(X, Y) >= Y  because [14], by (Select) 
  14] Y >= Y  by (Meta) 

  15] filter#(F, cons(X, Y)) >= filter#(F, Y)  because [10], by (Star) 

  16] if(X, Y) >= X  because [17], by (Star) 
  17] if*(X, Y) >= X  because [18], by (Select) 
  18] X >= X  by (Meta) 

  19] if(X, Y) >= Y  because [20], by (Star) 
  20] if*(X, Y) >= Y  because [21], by (Select) 
  21] Y >= Y  by (Meta) 

  22] sub(X) >= X  because [23], by (Star) 
  23] sub*(X) >= X  because [24], by (Select) 
  24] X >= X  by (Meta) 

  25] sub(s(X)) >= sub(X)  because [26], by (Star) 
  26] sub*(s(X)) >= sub(X)  because [27], by (Select) 
  27] s(X) >= sub(X)  because [28], by (Star) 
  28] s*(X) >= sub(X)  because s > sub and [29], by (Copy) 
  29] s*(X) >= X  because [30], by (Select) 
  30] X >= X  by (Meta) 

  31] gtr >= _|_  by (Bot) 

  32] gtr >= _|_  by (Bot) 

  33] gtr >= gtr  because gtr in Mul, by (Fun) 

  34] d(X, _|_) >= _|_  by (Bot) 

  35] d(s(X), s(Y)) >= if(_|_, d(s(X), sub(Y)))  because [36], by (Star) 
  36] d*(s(X), s(Y)) >= if(_|_, d(s(X), sub(Y)))  because d > if, [37] and [38], by (Copy) 
  37] d*(s(X), s(Y)) >= _|_  by (Bot) 
  38] d*(s(X), s(Y)) >= d(s(X), sub(Y))  because d in Mul, [39] and [41], by (Stat) 
  39] s(X) >= s(X)  because s in Mul and [40], by (Fun) 
  40] X >= X  by (Meta) 
  41] s(Y) > sub(Y)  because [42], by definition 
  42] s*(Y) >= sub(Y)  because s > sub and [43], by (Copy) 
  43] s*(Y) >= Y  because [44], by (Select) 
  44] Y >= Y  by (Meta) 

  45] len(_|_) >= _|_  by (Bot) 

  46] len(cons(X, Y)) >= s(len(Y))  because [47], by (Star) 
  47] len*(cons(X, Y)) >= s(len(Y))  because len = s, len in Mul and [48], by (Stat) 
  48] cons(X, Y) > len(Y)  because [49], by definition 
  49] cons*(X, Y) >= len(Y)  because cons > len and [50], by (Copy) 
  50] cons*(X, Y) >= Y  because [51], by (Select) 
  51] Y >= Y  by (Meta) 

  52] filter(F, _|_) >= _|_  by (Bot) 

  53] filter(F, cons(X, Y)) >= if(cons(X, filter(F, Y)), filter(F, Y))  because [54], by (Star) 
  54] filter*(F, cons(X, Y)) >= if(cons(X, filter(F, Y)), filter(F, Y))  because filter > if, [55] and [57], by (Copy) 
  55] filter*(F, cons(X, Y)) >= cons(X, filter(F, Y))  because filter > cons, [56] and [57], by (Copy) 
  56] filter*(F, cons(X, Y)) >= X  because [6], by (Select) 
  57] filter*(F, cons(X, Y)) >= filter(F, Y)  because filter in Mul, [11] and [12], by (Stat) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_2, R_0, minimal, formative), where P_2 consists of:

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

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 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)) >? F(X) 
  filter#(F, cons(X, Y)) >? filter#(F, Y) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  sub(X, 0) >= X 
  sub(s(X), s(Y)) >= sub(X, Y) 
  gtr(0, X) >= false 
  gtr(s(X), 0) >= true 
  gtr(s(X), s(Y)) >= gtr(X, Y) 
  d(X, 0) >= true 
  d(s(X), s(Y)) >= if(gtr(X, Y), false, d(s(X), sub(Y, X))) 
  len(nil) >= 0 
  len(cons(X, Y)) >= s(len(Y)) 
  filter(F, nil) >= nil 
  filter(F, cons(X, Y)) >= if(F X, cons(X, filter(F, Y)), filter(F, Y)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[0]] = _|_ 
  [[if(x_1, x_2, x_3)]] = if(x_2, x_3) 
  [[nil]] = _|_ 
  [[sub(x_1, x_2)]] = sub(x_1) 
  [[true]] = _|_ 

We choose Lex = {} and Mul = {@_{o -> o}, cons, d, false, filter, filter#, gtr, if, len, s, sub}, and the following precedence: filter# > gtr > filter > cons > @_{o -> o} > d > if > len > false > s > sub

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  filter#(F, cons(X, Y)) > @_{o -> o}(F, X) 
  filter#(F, cons(X, Y)) >= filter#(F, Y) 
  if(X, Y) >= X 
  if(X, Y) >= Y 
  sub(X) >= X 
  sub(s(X)) >= sub(X) 
  gtr(_|_, X) >= false 
  gtr(s(X), _|_) >= _|_ 
  gtr(s(X), s(Y)) >= gtr(X, Y) 
  d(X, _|_) >= _|_ 
  d(s(X), s(Y)) >= if(false, d(s(X), sub(Y))) 
  len(_|_) >= _|_ 
  len(cons(X, Y)) >= s(len(Y)) 
  filter(F, _|_) >= _|_ 
  filter(F, cons(X, Y)) >= if(cons(X, filter(F, Y)), filter(F, Y)) 

With these choices, we have:

  1] filter#(F, cons(X, Y)) > @_{o -> o}(F, X)  because [2], by definition 
  2] filter#*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter# > @_{o -> o}, [3] and [5], by (Copy) 
  3] filter#*(F, cons(X, Y)) >= F  because [4], by (Select) 
  4] F >= F  by (Meta) 
  5] filter#*(F, cons(X, Y)) >= X  because [6], by (Select) 
  6] cons(X, Y) >= X  because [7], by (Star) 
  7] cons*(X, Y) >= X  because [8], by (Select) 
  8] X >= X  by (Meta) 

  9] filter#(F, cons(X, Y)) >= filter#(F, Y)  because filter# in Mul, [10] and [11], by (Fun) 
  10] F >= F  by (Meta) 
  11] cons(X, Y) >= Y  because [12], by (Star) 
  12] cons*(X, Y) >= Y  because [13], by (Select) 
  13] Y >= Y  by (Meta) 

  14] if(X, Y) >= X  because [15], by (Star) 
  15] if*(X, Y) >= X  because [16], by (Select) 
  16] X >= X  by (Meta) 

  17] if(X, Y) >= Y  because [18], by (Star) 
  18] if*(X, Y) >= Y  because [19], by (Select) 
  19] Y >= Y  by (Meta) 

  20] sub(X) >= X  because [21], by (Star) 
  21] sub*(X) >= X  because [22], by (Select) 
  22] X >= X  by (Meta) 

  23] sub(s(X)) >= sub(X)  because [24], by (Star) 
  24] sub*(s(X)) >= sub(X)  because [25], by (Select) 
  25] s(X) >= sub(X)  because [26], by (Star) 
  26] s*(X) >= sub(X)  because s > sub and [27], by (Copy) 
  27] s*(X) >= X  because [28], by (Select) 
  28] X >= X  by (Meta) 

  29] gtr(_|_, X) >= false  because [30], by (Star) 
  30] gtr*(_|_, X) >= false  because gtr > false, by (Copy) 

  31] gtr(s(X), _|_) >= _|_  by (Bot) 

  32] gtr(s(X), s(Y)) >= gtr(X, Y)  because [33], by (Star) 
  33] gtr*(s(X), s(Y)) >= gtr(X, Y)  because gtr in Mul, [34] and [37], by (Stat) 
  34] s(X) > X  because [35], by definition 
  35] s*(X) >= X  because [36], by (Select) 
  36] X >= X  by (Meta) 
  37] s(Y) > Y  because [38], by definition 
  38] s*(Y) >= Y  because [39], by (Select) 
  39] Y >= Y  by (Meta) 

  40] d(X, _|_) >= _|_  by (Bot) 

  41] d(s(X), s(Y)) >= if(false, d(s(X), sub(Y)))  because [42], by (Star) 
  42] d*(s(X), s(Y)) >= if(false, d(s(X), sub(Y)))  because d > if, [43] and [44], by (Copy) 
  43] d*(s(X), s(Y)) >= false  because d > false, by (Copy) 
  44] d*(s(X), s(Y)) >= d(s(X), sub(Y))  because d in Mul, [45] and [47], by (Stat) 
  45] s(X) >= s(X)  because s in Mul and [46], by (Fun) 
  46] X >= X  by (Meta) 
  47] s(Y) > sub(Y)  because [48], by definition 
  48] s*(Y) >= sub(Y)  because s > sub and [49], by (Copy) 
  49] s*(Y) >= Y  because [50], by (Select) 
  50] Y >= Y  by (Meta) 

  51] len(_|_) >= _|_  by (Bot) 

  52] len(cons(X, Y)) >= s(len(Y))  because [53], by (Star) 
  53] len*(cons(X, Y)) >= s(len(Y))  because len > s and [54], by (Copy) 
  54] len*(cons(X, Y)) >= len(Y)  because len in Mul and [55], by (Stat) 
  55] cons(X, Y) > Y  because [56], by definition 
  56] cons*(X, Y) >= Y  because [57], by (Select) 
  57] Y >= Y  by (Meta) 

  58] filter(F, _|_) >= _|_  by (Bot) 

  59] filter(F, cons(X, Y)) >= if(cons(X, filter(F, Y)), filter(F, Y))  because [60], by (Star) 
  60] filter*(F, cons(X, Y)) >= if(cons(X, filter(F, Y)), filter(F, Y))  because filter > if, [61] and [63], by (Copy) 
  61] filter*(F, cons(X, Y)) >= cons(X, filter(F, Y))  because filter > cons, [62] and [63], by (Copy) 
  62] filter*(F, cons(X, Y)) >= X  because [6], by (Select) 
  63] filter*(F, cons(X, Y)) >= filter(F, Y)  because filter in Mul, [10] and [64], by (Stat) 
  64] cons(X, Y) > Y  because [65], by definition 
  65] cons*(X, Y) >= Y  because [13], by (Select) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_0, minimal, formative) by (P_3, R_0, minimal, formative), where P_3 consists of:

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

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

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

We apply the subterm criterion with the following projection function:

  nu(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_3, 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.