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TRS Stand 20472 pair #381710801
details
property
value
status
complete
benchmark
IJCAR_12.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n113.star.cs.uiowa.edu
space
AProVE_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.93821287155 seconds
cpu usage
4.482646851
max memory
2.2335488E8
stage attributes
key
value
output-size
10657
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 1 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES (15) QDP (16) QDPOrderProof [EQUIVALENT, 18 ms] (17) QDP (18) QDPOrderProof [EQUIVALENT, 0 ms] (19) QDP (20) TransformationProof [EQUIVALENT, 0 ms] (21) QDP (22) DependencyGraphProof [EQUIVALENT, 0 ms] (23) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: plus(x, 0) -> x plus(0, y) -> y plus(s(x), y) -> s(plus(x, y)) times(0, y) -> 0 times(s(0), y) -> y times(s(x), y) -> plus(y, times(x, y)) div(0, y) -> 0 div(x, y) -> quot(x, y, y) quot(0, s(y), z) -> 0 quot(s(x), s(y), z) -> quot(x, y, z) quot(x, 0, s(z)) -> s(div(x, s(z))) div(div(x, y), z) -> div(x, times(y, z)) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: PLUS(s(x), y) -> PLUS(x, y) TIMES(s(x), y) -> PLUS(y, times(x, y)) TIMES(s(x), y) -> TIMES(x, y) DIV(x, y) -> QUOT(x, y, y) QUOT(s(x), s(y), z) -> QUOT(x, y, z) QUOT(x, 0, s(z)) -> DIV(x, s(z)) DIV(div(x, y), z) -> DIV(x, times(y, z)) DIV(div(x, y), z) -> TIMES(y, z) The TRS R consists of the following rules: plus(x, 0) -> x plus(0, y) -> y plus(s(x), y) -> s(plus(x, y)) times(0, y) -> 0 times(s(0), y) -> y times(s(x), y) -> plus(y, times(x, y)) div(0, y) -> 0 div(x, y) -> quot(x, y, y) quot(0, s(y), z) -> 0 quot(s(x), s(y), z) -> quot(x, y, z) quot(x, 0, s(z)) -> s(div(x, s(z))) div(div(x, y), z) -> div(x, times(y, z)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ----------------------------------------
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