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HRS union beta 16688 pair #381734786
details
property
value
status
complete
benchmark
Applicative_AG01_innermost__#4.17.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n024.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
2.52729988098 seconds
cpu usage
4.518424432
max memory
1.41991936E8
stage attributes
key
value
output-size
2566
starexec-result
NO
output
/export/starexec/sandbox/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- 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(g(s(0)), 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(g(s(0)), %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(g(s(0)), %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 ||
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