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TRS Stand 20472 pair #381710966
details
property
value
status
complete
benchmark
rta1.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n021.star.cs.uiowa.edu
space
AProVE_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.67705202103 seconds
cpu usage
3.847878519
max memory
2.35761664E8
stage attributes
key
value
output-size
8073
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 17 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) MNOCProof [EQUIVALENT, 0 ms] (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) MRRProof [EQUIVALENT, 0 ms] (13) QDP (14) PisEmptyProof [EQUIVALENT, 0 ms] (15) YES (16) QDP (17) MNOCProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: plus(s(s(x)), y) -> s(plus(x, s(y))) plus(x, s(s(y))) -> s(plus(s(x), y)) plus(s(0), y) -> s(y) plus(0, y) -> y ack(0, y) -> s(y) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, plus(y, ack(s(x), y))) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: PLUS(s(s(x)), y) -> PLUS(x, s(y)) PLUS(x, s(s(y))) -> PLUS(s(x), y) ACK(s(x), 0) -> ACK(x, s(0)) ACK(s(x), s(y)) -> ACK(x, plus(y, ack(s(x), y))) ACK(s(x), s(y)) -> PLUS(y, ack(s(x), y)) ACK(s(x), s(y)) -> ACK(s(x), y) The TRS R consists of the following rules: plus(s(s(x)), y) -> s(plus(x, s(y))) plus(x, s(s(y))) -> s(plus(s(x), y)) plus(s(0), y) -> s(y) plus(0, y) -> y ack(0, y) -> s(y) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, plus(y, ack(s(x), y))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: PLUS(x, s(s(y))) -> PLUS(s(x), y)
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