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SRS Standard pair #487089424
details
property
value
status
complete
benchmark
abaaaa-aaaaabaabab.srs.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n147.star.cs.uiowa.edu
space
Wenzel_16
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.72664 seconds
cpu usage
3.96867
user time
3.79031
system time
0.178359
max virtual memory
1.9009336E7
max residence set size
339828.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Strip Symbols Proof [SOUND, 0 ms] (4) QTRS (5) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(a(a(a(x1)))))) -> a(a(a(a(a(b(a(a(b(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(a(a(a(b(a(x1)))))) -> b(a(b(a(a(b(a(a(a(a(a(x1))))))))))) Q is empty. ---------------------------------------- (3) Strip Symbols Proof (SOUND) We were given the following TRS: a(a(a(a(b(a(x1)))))) -> b(a(b(a(a(b(a(a(a(a(a(x1))))))))))) By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: a(a(a(a(b(x))))) -> b(a(b(a(a(b(a(a(a(a(x)))))))))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(a(b(x))))) -> b(a(b(a(a(b(a(a(a(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(a(a(a(b(x))))) -> b(a(b(a(a(b(a(a(a(a(x)))))))))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 489, 490, 491, 492, 493, 494, 495, 496, 497, 498, 499, 500, 501, 502, 503, 504, 505, 506, 507, 508 Node 489 is start node and node 490 is final node. Those nodes are connected through the following edges: * 489 to 491 labelled b_1(0)* 490 to 490 labelled #_1(0)* 491 to 492 labelled a_1(0)* 492 to 493 labelled b_1(0)* 493 to 494 labelled a_1(0)* 494 to 495 labelled a_1(0)* 495 to 496 labelled b_1(0)* 496 to 497 labelled a_1(0)* 496 to 500 labelled b_1(1)* 497 to 498 labelled a_1(0)* 497 to 500 labelled b_1(1)* 498 to 499 labelled a_1(0)* 498 to 500 labelled b_1(1)* 499 to 490 labelled a_1(0)* 499 to 500 labelled b_1(1)* 500 to 501 labelled a_1(1)* 501 to 502 labelled b_1(1)* 502 to 503 labelled a_1(1)* 503 to 504 labelled a_1(1)* 504 to 505 labelled b_1(1)* 505 to 506 labelled a_1(1)* 505 to 500 labelled b_1(1)* 506 to 507 labelled a_1(1)* 506 to 500 labelled b_1(1)* 507 to 508 labelled a_1(1)* 507 to 500 labelled b_1(1)* 508 to 490 labelled a_1(1)* 508 to 500 labelled b_1(1) ---------------------------------------- (6) YES
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