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SRS Standard pair #487518989
details
property
value
status
complete
benchmark
abaababa-aabababaab.srs.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n034.star.cs.uiowa.edu
space
Wenzel_16
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.61941599846 seconds
cpu usage
3.756026622
max memory
2.82443776E8
stage attributes
key
value
output-size
3063
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Strip Symbols Proof [SOUND, 0 ms] (4) QTRS (5) RFCMatchBoundsTRSProof [EQUIVALENT, 3 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(a(b(a(b(a(x1)))))))) -> a(a(b(a(b(a(b(a(a(b(x1)))))))))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(b(a(a(b(a(x1)))))))) -> b(a(a(b(a(b(a(b(a(a(x1)))))))))) Q is empty. ---------------------------------------- (3) Strip Symbols Proof (SOUND) We were given the following TRS: a(b(a(b(a(a(b(a(x1)))))))) -> b(a(a(b(a(b(a(b(a(a(x1)))))))))) By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: a(b(a(b(a(a(b(x))))))) -> b(a(a(b(a(b(a(b(a(x))))))))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(b(a(a(b(x))))))) -> b(a(a(b(a(b(a(b(a(x))))))))) Q is empty. ---------------------------------------- (5) 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: a(b(a(b(a(a(b(x))))))) -> b(a(a(b(a(b(a(b(a(x))))))))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 87, 88, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 116, 118, 120, 122, 124, 126 Node 87 is start node and node 88 is final node. Those nodes are connected through the following edges: * 87 to 105 labelled b_1(0)* 88 to 88 labelled #_1(0)* 105 to 106 labelled a_1(0)* 106 to 107 labelled a_1(0)* 107 to 108 labelled b_1(0)* 108 to 109 labelled a_1(0)* 108 to 113 labelled b_1(1)* 109 to 110 labelled b_1(0)* 110 to 111 labelled a_1(0)* 110 to 113 labelled b_1(1)* 111 to 112 labelled b_1(0)* 112 to 88 labelled a_1(0)* 112 to 113 labelled b_1(1)* 113 to 114 labelled a_1(1)* 114 to 116 labelled a_1(1)* 116 to 118 labelled b_1(1)* 118 to 120 labelled a_1(1)* 118 to 113 labelled b_1(1)* 120 to 122 labelled b_1(1)* 122 to 124 labelled a_1(1)* 122 to 113 labelled b_1(1)* 124 to 126 labelled b_1(1)* 126 to 88 labelled a_1(1)* 126 to 113 labelled b_1(1) ---------------------------------------- (6) YES
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