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SRS Stand 10685 pair #381718414
details
property
value
status
complete
benchmark
aprove01.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n038.star.cs.uiowa.edu
space
Secret_06_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
4.43488383293 seconds
cpu usage
13.402412264
max memory
1.183260672E9
stage attributes
key
value
output-size
18599
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 23 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) MNOCProof [EQUIVALENT, 0 ms] (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] (18) YES (19) QDP (20) MNOCProof [EQUIVALENT, 0 ms] (21) QDP (22) UsableRulesProof [EQUIVALENT, 0 ms] (23) QDP (24) QReductionProof [EQUIVALENT, 1 ms] (25) QDP (26) QDPOrderProof [EQUIVALENT, 16 ms] (27) QDP (28) QDPOrderProof [EQUIVALENT, 17 ms] (29) QDP (30) PisEmptyProof [EQUIVALENT, 0 ms] (31) YES (32) QDP (33) MNOCProof [EQUIVALENT, 0 ms] (34) QDP (35) UsableRulesProof [EQUIVALENT, 0 ms] (36) QDP (37) QReductionProof [EQUIVALENT, 0 ms] (38) QDP (39) QDPOrderProof [EQUIVALENT, 201 ms] (40) QDP (41) DependencyGraphProof [EQUIVALENT, 0 ms] (42) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(0(x1)) -> 0(s(s(p(x1)))) p(s(x1)) -> x1 p(p(s(x1))) -> p(x1) f(s(x1)) -> p(s(g(p(s(s(x1)))))) g(s(x1)) -> p(p(s(s(s(j(s(p(s(p(s(x1))))))))))) j(s(x1)) -> p(s(s(p(s(f(p(s(p(p(s(x1))))))))))) half(0(x1)) -> 0(s(s(half(p(s(p(s(x1)))))))) half(s(s(x1))) -> s(half(p(p(s(s(x1)))))) rd(0(x1)) -> 0(s(0(0(0(0(s(0(rd(x1))))))))) 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: P(0(x1)) -> P(x1) P(p(s(x1))) -> P(x1) F(s(x1)) -> P(s(g(p(s(s(x1)))))) F(s(x1)) -> G(p(s(s(x1)))) F(s(x1)) -> P(s(s(x1))) G(s(x1)) -> P(p(s(s(s(j(s(p(s(p(s(x1))))))))))) G(s(x1)) -> P(s(s(s(j(s(p(s(p(s(x1)))))))))) G(s(x1)) -> J(s(p(s(p(s(x1)))))) G(s(x1)) -> P(s(p(s(x1)))) G(s(x1)) -> P(s(x1)) J(s(x1)) -> P(s(s(p(s(f(p(s(p(p(s(x1))))))))))) J(s(x1)) -> P(s(f(p(s(p(p(s(x1))))))))
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