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TRS Stand 20472 pair #381713794
details
property
value
status
complete
benchmark
ExAppendixB_AEL03_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n192.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.8370821476 seconds
cpu usage
4.070617905
max memory
2.5042944E8
stage attributes
key
value
output-size
9493
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 4 ms] (2) QDP (3) DependencyGraphProof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) QDPSizeChangeProof [EQUIVALENT, 0 ms] (17) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: from(X) -> cons(X, n__from(s(X))) 2ndspos(0, Z) -> rnil 2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z))) 2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z))) 2ndsneg(0, Z) -> rnil 2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z))) 2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, activate(Z))) pi(X) -> 2ndspos(X, from(0)) plus(0, Y) -> Y plus(s(X), Y) -> s(plus(X, Y)) times(0, Y) -> 0 times(s(X), Y) -> plus(Y, times(X, Y)) square(X) -> times(X, X) from(X) -> n__from(X) activate(n__from(X)) -> from(X) activate(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: 2NDSPOS(s(N), cons(X, Z)) -> 2NDSPOS(s(N), cons2(X, activate(Z))) 2NDSPOS(s(N), cons(X, Z)) -> ACTIVATE(Z) 2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, activate(Z)) 2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> ACTIVATE(Z) 2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, activate(Z))) 2NDSNEG(s(N), cons(X, Z)) -> ACTIVATE(Z) 2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, activate(Z)) 2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> ACTIVATE(Z) PI(X) -> 2NDSPOS(X, from(0)) PI(X) -> FROM(0) PLUS(s(X), Y) -> PLUS(X, Y) TIMES(s(X), Y) -> PLUS(Y, times(X, Y)) TIMES(s(X), Y) -> TIMES(X, Y) SQUARE(X) -> TIMES(X, X) ACTIVATE(n__from(X)) -> FROM(X) The TRS R consists of the following rules: from(X) -> cons(X, n__from(s(X))) 2ndspos(0, Z) -> rnil 2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z))) 2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z))) 2ndsneg(0, Z) -> rnil 2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z))) 2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, activate(Z))) pi(X) -> 2ndspos(X, from(0)) plus(0, Y) -> Y plus(s(X), Y) -> s(plus(X, Y)) times(0, Y) -> 0 times(s(X), Y) -> plus(Y, times(X, Y))
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