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TRS Standard pair #516968130
details
property
value
status
complete
benchmark
t004.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n029.star.cs.uiowa.edu
space
HirokawaMiddeldorp_04
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.08581280708 seconds
cpu usage
5.340410536
max memory
4.71105536E8
stage attributes
key
value
output-size
2329
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms] (4)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: f(s(x)) -> s(f(f(p(s(x))))) f(0) -> 0 p(s(x)) -> 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: s(f(x)) -> s(p(f(f(s(x))))) 0'(f(x)) -> 0'(x) s(p(x)) -> x Q is empty. ---------------------------------------- (3) RFCMatchBoundsTRSProof (EQUIVALENT) Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 2. This implies Q-termination of R. The following rules were used to construct the certificate: s(f(x)) -> s(p(f(f(s(x))))) 0'(f(x)) -> 0'(x) s(p(x)) -> x The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 3, 4, 13, 14, 15, 16, 17, 18, 19, 20 Node 3 is start node and node 4 is final node. Those nodes are connected through the following edges: * 3 to 13 labelled s_1(0)* 3 to 4 labelled 0'_1(0), s_1(0), f_1(0), p_1(0), 0'_1(1), s_1(1), f_1(1), p_1(1)* 3 to 17 labelled s_1(1)* 3 to 15 labelled f_1(1)* 3 to 19 labelled f_1(2)* 4 to 4 labelled #_1(0)* 13 to 14 labelled p_1(0)* 14 to 15 labelled f_1(0)* 15 to 16 labelled f_1(0)* 16 to 4 labelled s_1(0), s_1(1), f_1(1), p_1(1), 0'_1(1)* 16 to 17 labelled s_1(1)* 16 to 19 labelled f_1(2)* 17 to 18 labelled p_1(1)* 18 to 19 labelled f_1(1)* 19 to 20 labelled f_1(1)* 20 to 4 labelled s_1(1), f_1(1), p_1(1), 0'_1(1)* 20 to 17 labelled s_1(1)* 20 to 19 labelled f_1(2) ---------------------------------------- (4) YES
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