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

  Rules:

    f(g(x), s(0), y) => f(y, y, g(x)) 
    g(s(x)) => s(g(x)) 
    g(0) => 0 
    map(h, nil) => nil 
    map(h, cons(x, y)) => cons(h x, map(h, y)) 
    filter(h, nil) => nil 
    filter(h, cons(x, y)) => filter2(h x, h, x, y) 
    filter2(true, h, x, y) => cons(x, filter(h, y)) 
    filter2(false, h, x, y) => filter(h, 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(s(0)), g(s(0)), g(s(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(g(%X), s(0), %Y) -> f(%Y, %Y, g(%X))
 ||    g(s(%X)) -> s(g(%X))
 ||    g(0) -> 0
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) NonTerminationProof (EQUIVALENT)
 || The following loops were found:
 || 
 || ---------- Loop: ----------
 || 
 || f(g(s(0)), g(s(0)), g(s(0))) -> f(g(s(0)), s(g(0)), g(s(0))) with rule g(s(%X')) -> s(g(%X')) at position [1] and matcher [%X' / 0]
 || 
 || f(g(s(0)), s(g(0)), g(s(0))) -> f(g(s(0)), s(0), g(s(0))) with rule g(0) -> 0 at position [1,0] and matcher [ ]
 || 
 || f(g(s(0)), s(0), g(s(0))) -> f(g(s(0)), g(s(0)), g(s(0))) with rule f(g(%X), s(0), %Y) -> f(%Y, %Y, g(%X)) at position [] and matcher [%X / s(0), %Y / g(s(0))]
 || 
 || Now an instance of the first term with Matcher [ ] occurs in the last term at position [].
 || 
 || Context: []
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || NO
 ||