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SRS Standard pair #487519865
details
property
value
status
complete
benchmark
z050.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n185.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.69542598724 seconds
cpu usage
3.615293257
max memory
2.54316544E8
stage attributes
key
value
output-size
2060
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Strip Symbols Proof [SOUND, 0 ms] (2) QTRS (3) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(b(a(a(b(x1)))))) -> a(a(b(b(a(b(a(x1))))))) Q is empty. ---------------------------------------- (1) Strip Symbols Proof (SOUND) We were given the following TRS: a(b(b(a(a(b(x1)))))) -> a(a(b(b(a(b(a(x1))))))) By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: b(b(a(a(b(x1))))) -> a(b(b(a(b(a(x1)))))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(a(a(b(x1))))) -> a(b(b(a(b(a(x1)))))) Q is empty. ---------------------------------------- (3) RFCMatchBoundsTRSProof (EQUIVALENT) Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 0. This implies Q-termination of R. The following rules were used to construct the certificate: b(b(a(a(b(x1))))) -> a(b(b(a(b(a(x1)))))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 3, 4, 11, 12, 13, 14, 15 Node 3 is start node and node 4 is final node. Those nodes are connected through the following edges: * 3 to 11 labelled a_1(0)* 4 to 4 labelled #_1(0)* 11 to 12 labelled b_1(0)* 12 to 13 labelled b_1(0)* 13 to 14 labelled a_1(0)* 14 to 15 labelled b_1(0)* 15 to 4 labelled a_1(0) ---------------------------------------- (4) YES
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