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TRS Stand 20472 pair #381713742
details
property
value
status
complete
benchmark
#3.7.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n010.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.72199296951 seconds
cpu usage
3.603388658
max memory
2.46411264E8
stage attributes
key
value
output-size
2479
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: half(0) -> 0 half(s(s(x))) -> s(half(x)) log(s(0)) -> 0 log(s(s(x))) -> s(log(s(half(x)))) 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: 0'(half(x)) -> 0'(x) s(s(half(x))) -> half(s(x)) 0'(s(log(x))) -> 0'(x) s(s(log(x))) -> half(s(log(s(x)))) 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: 0'(half(x)) -> 0'(x) s(s(half(x))) -> half(s(x)) 0'(s(log(x))) -> 0'(x) s(s(log(x))) -> half(s(log(s(x)))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 28, 29, 30, 31, 32, 33, 34, 35, 36, 37 Node 28 is start node and node 29 is final node. Those nodes are connected through the following edges: * 28 to 29 labelled 0'_1(0), 0'_1(1)* 28 to 30 labelled half_1(0)* 28 to 31 labelled half_1(0)* 29 to 29 labelled #_1(0)* 30 to 29 labelled s_1(0)* 30 to 34 labelled half_1(1)* 30 to 35 labelled half_1(1)* 31 to 32 labelled s_1(0)* 32 to 33 labelled log_1(0)* 33 to 29 labelled s_1(0)* 33 to 34 labelled half_1(1)* 33 to 35 labelled half_1(1)* 34 to 29 labelled s_1(1)* 34 to 34 labelled half_1(1)* 34 to 35 labelled half_1(1)* 35 to 36 labelled s_1(1)* 36 to 37 labelled log_1(1)* 37 to 29 labelled s_1(1)* 37 to 34 labelled half_1(1)* 37 to 35 labelled half_1(1) ---------------------------------------- (4) YES
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