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SRS Stand 10685 pair #381721958
details
property
value
status
complete
benchmark
baaabbaaab-aaabbaaabaaabb.srs.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n012.star.cs.uiowa.edu
space
Wenzel_16
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.57159590721 seconds
cpu usage
3.53752092
max memory
2.6185728E8
stage attributes
key
value
output-size
2993
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Strip Symbols Proof [SOUND, 0 ms] (2) QTRS (3) RFCMatchBoundsTRSProof [EQUIVALENT, 2 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(a(a(b(b(a(a(a(b(x1)))))))))) -> a(a(a(b(b(a(a(a(b(a(a(a(b(b(x1)))))))))))))) Q is empty. ---------------------------------------- (1) Strip Symbols Proof (SOUND) We were given the following TRS: b(a(a(a(b(b(a(a(a(b(x1)))))))))) -> a(a(a(b(b(a(a(a(b(a(a(a(b(b(x1)))))))))))))) By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: b(a(a(a(b(b(a(a(a(x))))))))) -> a(a(a(b(b(a(a(a(b(a(a(a(b(x))))))))))))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(a(a(b(b(a(a(a(x))))))))) -> a(a(a(b(b(a(a(a(b(a(a(a(b(x))))))))))))) Q is empty. ---------------------------------------- (3) RFCMatchBoundsTRSProof (EQUIVALENT) Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1. This implies Q-termination of R. The following rules were used to construct the certificate: b(a(a(a(b(b(a(a(a(x))))))))) -> a(a(a(b(b(a(a(a(b(a(a(a(b(x))))))))))))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 109, 110, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160 Node 109 is start node and node 110 is final node. Those nodes are connected through the following edges: * 109 to 137 labelled a_1(0)* 110 to 110 labelled #_1(0)* 137 to 138 labelled a_1(0)* 138 to 139 labelled a_1(0)* 139 to 140 labelled b_1(0)* 140 to 141 labelled b_1(0)* 141 to 142 labelled a_1(0)* 142 to 143 labelled a_1(0)* 143 to 144 labelled a_1(0)* 144 to 145 labelled b_1(0)* 144 to 149 labelled a_1(1)* 145 to 146 labelled a_1(0)* 146 to 147 labelled a_1(0)* 147 to 148 labelled a_1(0)* 148 to 110 labelled b_1(0)* 148 to 149 labelled a_1(1)* 149 to 150 labelled a_1(1)* 150 to 151 labelled a_1(1)* 151 to 152 labelled b_1(1)* 152 to 153 labelled b_1(1)* 153 to 154 labelled a_1(1)* 154 to 155 labelled a_1(1)* 155 to 156 labelled a_1(1)* 156 to 157 labelled b_1(1)* 156 to 149 labelled a_1(1)* 157 to 158 labelled a_1(1)* 158 to 159 labelled a_1(1)* 159 to 160 labelled a_1(1)* 160 to 110 labelled b_1(1)* 160 to 149 labelled a_1(1) ---------------------------------------- (4) YES
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