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TRS Stand 20472 pair #381715722
details
property
value
status
complete
benchmark
Ex4_7_37_Bor03_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n089.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.79595708847 seconds
cpu usage
4.101020776
max memory
2.56811008E8
stage attributes
key
value
output-size
11265
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDPOrderProof [EQUIVALENT, 14 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES (15) QDP (16) UsableRulesProof [EQUIVALENT, 0 ms] (17) QDP (18) QDPSizeChangeProof [EQUIVALENT, 0 ms] (19) YES (20) QDP (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: from(X) -> cons(X, n__from(s(X))) sel(0, cons(X, XS)) -> X sel(s(N), cons(X, XS)) -> sel(N, activate(XS)) minus(X, 0) -> 0 minus(s(X), s(Y)) -> minus(X, Y) quot(0, s(Y)) -> 0 quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y))) zWquot(XS, nil) -> nil zWquot(nil, XS) -> nil zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), n__zWquot(activate(XS), activate(YS))) from(X) -> n__from(X) zWquot(X1, X2) -> n__zWquot(X1, X2) activate(n__from(X)) -> from(X) activate(n__zWquot(X1, X2)) -> zWquot(X1, X2) 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: SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS)) SEL(s(N), cons(X, XS)) -> ACTIVATE(XS) MINUS(s(X), s(Y)) -> MINUS(X, Y) QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y)) QUOT(s(X), s(Y)) -> MINUS(X, Y) ZWQUOT(cons(X, XS), cons(Y, YS)) -> QUOT(X, Y) ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS) ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS) ACTIVATE(n__from(X)) -> FROM(X) ACTIVATE(n__zWquot(X1, X2)) -> ZWQUOT(X1, X2) The TRS R consists of the following rules: from(X) -> cons(X, n__from(s(X))) sel(0, cons(X, XS)) -> X sel(s(N), cons(X, XS)) -> sel(N, activate(XS)) minus(X, 0) -> 0 minus(s(X), s(Y)) -> minus(X, Y) quot(0, s(Y)) -> 0 quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y))) zWquot(XS, nil) -> nil zWquot(nil, XS) -> nil zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), n__zWquot(activate(XS), activate(YS))) from(X) -> n__from(X) zWquot(X1, X2) -> n__zWquot(X1, X2) activate(n__from(X)) -> from(X)
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