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TRS Standard pair #516960800
details
property
value
status
complete
benchmark
kabasci04.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n048.star.cs.uiowa.edu
space
AProVE_07
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.40769386292 seconds
cpu usage
6.189011192
max memory
4.80063488E8
stage attributes
key
value
output-size
11608
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 23 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES (15) QDP (16) UsableRulesProof [EQUIVALENT, 0 ms] (17) QDP (18) QDPSizeChangeProof [EQUIVALENT, 0 ms] (19) YES (20) QDP (21) UsableRulesProof [EQUIVALENT, 0 ms] (22) QDP (23) QDPSizeChangeProof [EQUIVALENT, 0 ms] (24) YES (25) QDP (26) QDPOrderProof [EQUIVALENT, 13 ms] (27) QDP (28) PisEmptyProof [EQUIVALENT, 0 ms] (29) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: min(x, 0) -> 0 min(0, y) -> 0 min(s(x), s(y)) -> s(min(x, y)) max(x, 0) -> x max(0, y) -> y max(s(x), s(y)) -> s(max(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> s(minus(x, any(y))) gcd(s(x), s(y)) -> gcd(minus(max(x, y), min(x, y)), s(min(x, y))) any(s(x)) -> s(s(any(x))) any(x) -> x 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: MIN(s(x), s(y)) -> MIN(x, y) MAX(s(x), s(y)) -> MAX(x, y) MINUS(s(x), s(y)) -> MINUS(x, any(y)) MINUS(s(x), s(y)) -> ANY(y) GCD(s(x), s(y)) -> GCD(minus(max(x, y), min(x, y)), s(min(x, y))) GCD(s(x), s(y)) -> MINUS(max(x, y), min(x, y)) GCD(s(x), s(y)) -> MAX(x, y) GCD(s(x), s(y)) -> MIN(x, y) ANY(s(x)) -> ANY(x) The TRS R consists of the following rules: min(x, 0) -> 0 min(0, y) -> 0 min(s(x), s(y)) -> s(min(x, y)) max(x, 0) -> x max(0, y) -> y max(s(x), s(y)) -> s(max(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> s(minus(x, any(y))) gcd(s(x), s(y)) -> gcd(minus(max(x, y), min(x, y)), s(min(x, y))) any(s(x)) -> s(s(any(x))) any(x) -> x
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