/export/starexec/sandbox/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

  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 interpretation [POLO]:
 || 
 ||    POL(0) = 0
 ||    POL(D(x_1, x_2)) = x_2
 ||    POL(s(x_1)) = 1 + x_1
 ||    POL(sub(x_1, x_2)) = x_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 static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]).

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)) =#> filter#(F, Y)   
    2] 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 
    * 2 : 0, 1, 2 

This graph has the following strongly connected components:

  P_1:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, 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 +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.