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TRS Stand 20472 pair #381715177
details
property
value
status
complete
benchmark
Ex49_GM04_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n082.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.92273712158 seconds
cpu usage
4.608191572
max memory
2.7277312E8
stage attributes
key
value
output-size
12750
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) MNOCProof [EQUIVALENT, 0 ms] (9) QDP (10) MRRProof [EQUIVALENT, 0 ms] (11) QDP (12) PisEmptyProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) MNOCProof [EQUIVALENT, 0 ms] (18) QDP (19) MRRProof [EQUIVALENT, 7 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES (23) QDP (24) QDPOrderProof [EQUIVALENT, 0 ms] (25) QDP (26) PisEmptyProof [EQUIVALENT, 0 ms] (27) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: minus(n__0, Y) -> 0 minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y)) geq(X, n__0) -> true geq(n__0, n__s(Y)) -> false geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y)) div(0, n__s(Y)) -> 0 div(s(X), n__s(Y)) -> if(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0) if(true, X, Y) -> activate(X) if(false, X, Y) -> activate(Y) 0 -> n__0 s(X) -> n__s(X) activate(n__0) -> 0 activate(n__s(X)) -> s(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: MINUS(n__0, Y) -> 0^1 MINUS(n__s(X), n__s(Y)) -> MINUS(activate(X), activate(Y)) MINUS(n__s(X), n__s(Y)) -> ACTIVATE(X) MINUS(n__s(X), n__s(Y)) -> ACTIVATE(Y) GEQ(n__s(X), n__s(Y)) -> GEQ(activate(X), activate(Y)) GEQ(n__s(X), n__s(Y)) -> ACTIVATE(X) GEQ(n__s(X), n__s(Y)) -> ACTIVATE(Y) DIV(s(X), n__s(Y)) -> IF(geq(X, activate(Y)), n__s(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0) DIV(s(X), n__s(Y)) -> GEQ(X, activate(Y)) DIV(s(X), n__s(Y)) -> ACTIVATE(Y) DIV(s(X), n__s(Y)) -> DIV(minus(X, activate(Y)), n__s(activate(Y))) DIV(s(X), n__s(Y)) -> MINUS(X, activate(Y)) IF(true, X, Y) -> ACTIVATE(X) IF(false, X, Y) -> ACTIVATE(Y) ACTIVATE(n__0) -> 0^1 ACTIVATE(n__s(X)) -> S(X) The TRS R consists of the following rules: minus(n__0, Y) -> 0 minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y)) geq(X, n__0) -> true
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