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


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NO
We consider the system theBenchmark.

  Alphabet:

    0 : [] --> b 
    1 : [] --> b 
    cons : [e * f] --> f 
    f : [b * b * b * c] --> a 
    false : [] --> d 
    filter : [e -> d * f] --> f 
    filter2 : [d * e -> d * e * f] --> f 
    g : [b * b] --> b 
    h : [b] --> c 
    map : [e -> e * f] --> f 
    nil : [] --> f 
    true : [] --> d 

  Rules:

    f(0, 1, g(x, y), z) => f(g(x, y), g(x, y), g(x, y), h(x)) 
    g(0, 1) => 0 
    g(0, 1) => 1 
    h(g(x, y)) => h(x) 
    map(i, nil) => nil 
    map(i, cons(x, y)) => cons(i x, map(i, y)) 
    filter(i, nil) => nil 
    filter(i, cons(x, y)) => filter2(i x, i, x, y) 
    filter2(true, i, x, y) => cons(x, filter(i, y)) 
    filter2(false, i, x, y) => filter(i, y) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

This system is non-terminating, as demonstrated by our external first-order non-termination checker:

 || The following well-typed term is terminating: f(g(0, 1), g(0, 1), g(0, 1), h(0))
 || 
 || proof of resources/system.trs
 || # AProVE Commit ID: 500ec9b2e2a919720cb177ef26031cb0220e008e fuhs 20130603
 || 
 || 
 || Termination w.r.t. Q of the given QTRS could be disproven:
 || 
 || (0) QTRS
 || (1) NonTerminationProof [EQUIVALENT, 0 ms]
 || (2) NO
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    f(0, 1, g(%X, %Y), %Z) -> f(g(%X, %Y), g(%X, %Y), g(%X, %Y), h(%X))
 ||    g(0, 1) -> 0
 ||    g(0, 1) -> 1
 ||    h(g(%X, %Y)) -> h(%X)
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) NonTerminationProof (EQUIVALENT)
 || The following loops were found:
 || 
 || ---------- Loop: ----------
 || 
 || f(g(0, 1), g(0, 1), g(0, 1), h(0)) -> f(g(0, 1), 1, g(0, 1), h(0)) with rule g(0, 1) -> 1 at position [1] and matcher [ ]
 || 
 || f(g(0, 1), 1, g(0, 1), h(0)) -> f(0, 1, g(0, 1), h(0)) with rule g(0, 1) -> 0 at position [0] and matcher [ ]
 || 
 || f(0, 1, g(0, 1), h(0)) -> f(g(0, 1), g(0, 1), g(0, 1), h(0)) with rule f(0, 1, g(%X, %Y), %Z) -> f(g(%X, %Y), g(%X, %Y), g(%X, %Y), h(%X)) at position [] and matcher [%X / 0, %Y / 1, %Z / h(0)]
 || 
 || Now an instance of the first term with Matcher [ ] occurs in the last term at position [].
 || 
 || Context: []
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || NO
 ||