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TRS Innermost pair #516979465
details
property
value
status
complete
benchmark
Ex15_Luc06_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n118.star.cs.uiowa.edu
space
Transformed_CSR_innermost_04
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.0205039978 seconds
cpu usage
5.433741248
max memory
4.72326144E8
stage attributes
key
value
output-size
2526
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [SOUND, 0 ms] (2) QTRS (3) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms] (4) YES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__f(f(a)) -> a__f(g(f(a))) mark(f(X)) -> a__f(X) mark(a) -> a mark(g(X)) -> g(mark(X)) a__f(X) -> f(X) The set Q consists of the following terms: mark(f(x0)) mark(a) mark(g(x0)) a__f(x0) ---------------------------------------- (1) QTRS Reverse (SOUND) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a'(f(a__f(x))) -> a'(f(g(a__f(x)))) f(mark(X)) -> a__f(X) a'(mark(x)) -> a'(x) g(mark(X)) -> mark(g(X)) a__f(X) -> f(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: a'(f(a__f(x))) -> a'(f(g(a__f(x)))) f(mark(X)) -> a__f(X) a'(mark(x)) -> a'(x) g(mark(X)) -> mark(g(X)) a__f(X) -> f(X) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 Node 11 is start node and node 12 is final node. Those nodes are connected through the following edges: * 11 to 13 labelled a'_1(0)* 11 to 12 labelled a__f_1(0), a'_1(0), f_1(0), f_1(1), a'_1(1), a__f_1(1), f_1(2)* 11 to 16 labelled mark_1(0)* 11 to 17 labelled a'_1(1)* 12 to 12 labelled #_1(0)* 13 to 14 labelled f_1(0)* 14 to 15 labelled g_1(0)* 15 to 12 labelled a__f_1(0), f_1(1), a__f_1(1), f_1(2)* 16 to 12 labelled g_1(0)* 16 to 20 labelled mark_1(1)* 17 to 18 labelled f_1(1)* 18 to 19 labelled g_1(1)* 19 to 12 labelled a__f_1(1), f_1(2)* 20 to 12 labelled g_1(1)* 20 to 20 labelled mark_1(1) ---------------------------------------- (4) YES
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