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SRS Standard pair #516968704
details
property
value
status
complete
benchmark
z056.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
Zantema_04
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.17399907112 seconds
cpu usage
5.682187293
max memory
5.0597888E8
stage attributes
key
value
output-size
2143
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms] (4)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: b(a(b(a(b(b(x1)))))) -> a(b(a(b(b(b(a(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: b(b(a(b(a(b(x1)))))) -> a(b(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 1. This implies Q-termination of R. The following rules were used to construct the certificate: b(b(a(b(a(b(x1)))))) -> a(b(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, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34 Node 3 is start node and node 4 is final node. Those nodes are connected through the following edges: * 3 to 23 labelled a_1(0)* 4 to 4 labelled #_1(0)* 23 to 24 labelled b_1(0)* 24 to 25 labelled b_1(0)* 24 to 29 labelled a_1(1)* 25 to 26 labelled b_1(0)* 26 to 27 labelled a_1(0)* 27 to 28 labelled b_1(0)* 28 to 4 labelled a_1(0)* 29 to 30 labelled b_1(1)* 30 to 31 labelled b_1(1)* 30 to 29 labelled a_1(1)* 31 to 32 labelled b_1(1)* 32 to 33 labelled a_1(1)* 33 to 34 labelled b_1(1)* 34 to 4 labelled a_1(1) ---------------------------------------- (4) YES
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