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SRS Standard pair #487090696
details
property
value
status
complete
benchmark
ababaaaaa-aaaaaababab.srs.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n054.star.cs.uiowa.edu
space
Wenzel_16
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.75844 seconds
cpu usage
3.99384
user time
3.8185
system time
0.175346
max virtual memory
1.907478E7
max residence set size
309120.0
stage attributes
key
value
starexec-result
YES
output
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Strip Symbols Proof [SOUND, 0 ms] (4)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, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(b(a(a(a(a(a(x1))))))))) -> a(a(a(a(a(a(b(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(a(b(a(b(a(x1))))))))) -> b(a(b(a(b(a(a(a(a(a(a(x1))))))))))) Q is empty. ---------------------------------------- (3) Strip Symbols Proof (SOUND) We were given the following TRS: a(a(a(a(a(b(a(b(a(x1))))))))) -> b(a(b(a(b(a(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(a(b(a(b(x)))))))) -> b(a(b(a(b(a(a(a(a(a(x)))))))))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(a(a(b(a(b(x)))))))) -> b(a(b(a(b(a(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(a(b(a(b(x)))))))) -> b(a(b(a(b(a(a(a(a(a(x)))))))))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237 Node 218 is start node and node 219 is final node. Those nodes are connected through the following edges: * 218 to 220 labelled b_1(0)* 219 to 219 labelled #_1(0)* 220 to 221 labelled a_1(0)* 221 to 222 labelled b_1(0)* 222 to 223 labelled a_1(0)* 223 to 224 labelled b_1(0)* 224 to 225 labelled a_1(0)* 224 to 229 labelled b_1(1)* 225 to 226 labelled a_1(0)* 225 to 229 labelled b_1(1)* 226 to 227 labelled a_1(0)* 226 to 229 labelled b_1(1)* 227 to 228 labelled a_1(0)* 227 to 229 labelled b_1(1)* 228 to 219 labelled a_1(0)* 228 to 229 labelled b_1(1)* 229 to 230 labelled a_1(1)* 230 to 231 labelled b_1(1)* 231 to 232 labelled a_1(1)* 232 to 233 labelled b_1(1)* 233 to 234 labelled a_1(1)* 233 to 229 labelled b_1(1)* 234 to 235 labelled a_1(1)* 234 to 229 labelled b_1(1)* 235 to 236 labelled a_1(1)* 235 to 229 labelled b_1(1)* 236 to 237 labelled a_1(1)* 236 to 229 labelled b_1(1)* 237 to 219 labelled a_1(1)* 237 to 229 labelled b_1(1) ---------------------------------------- (6) YES
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