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TRS Standard pair #487068153
details
property
value
status
complete
benchmark
4.43.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n175.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.78521 seconds
cpu usage
3.89081
user time
3.71612
system time
0.174689
max virtual memory
1.8408912E7
max residence set size
250312.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 52 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) RisEmptyProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: +(x, 0) -> x +(x, s(y)) -> s(+(x, y)) +(0, y) -> y +(s(x), y) -> s(+(x, y)) +(x, +(y, z)) -> +(+(x, y), z) f(g(f(x))) -> f(h(s(0), x)) f(g(h(x, y))) -> f(h(s(x), y)) f(h(x, h(y, z))) -> f(h(+(x, y), z)) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(+(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(0) = 0 POL(f(x_1)) = 2 + x_1 POL(g(x_1)) = 1 + 2*x_1 POL(h(x_1, x_2)) = 2 + x_1 + 2*x_2 POL(s(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: +(x, 0) -> x +(0, y) -> y +(x, +(y, z)) -> +(+(x, y), z) f(g(f(x))) -> f(h(s(0), x)) f(g(h(x, y))) -> f(h(s(x), y)) f(h(x, h(y, z))) -> f(h(+(x, y), z)) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: +(x, s(y)) -> s(+(x, y)) +(s(x), y) -> s(+(x, y)) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Knuth-Bendix order [KBO] with precedence:+_2 > s_1 and weight map: s_1=1 +_2=0 The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: +(x, s(y)) -> s(+(x, y)) +(s(x), y) -> s(+(x, y)) ---------------------------------------- (4) Obligation: Q restricted rewrite system: R is empty. Q is empty. ---------------------------------------- (5) RisEmptyProof (EQUIVALENT)
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