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TRS Stand 20472 pair #381715427
details
property
value
status
complete
benchmark
PEANO_complete-noand_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n092.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
5.58393001556 seconds
cpu usage
18.613303865
max memory
1.59440896E9
stage attributes
key
value
output-size
45736
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDP (5) QDPOrderProof [EQUIVALENT, 837 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) AND (9) QDP (10) QDPOrderProof [EQUIVALENT, 169 ms] (11) QDP (12) DependencyGraphProof [EQUIVALENT, 0 ms] (13) TRUE (14) QDP (15) QDPOrderProof [EQUIVALENT, 53 ms] (16) QDP (17) DependencyGraphProof [EQUIVALENT, 0 ms] (18) TRUE (19) QDP (20) UsableRulesProof [EQUIVALENT, 0 ms] (21) QDP (22) QDPSizeChangeProof [EQUIVALENT, 0 ms] (23) YES (24) QDP (25) QDPOrderProof [EQUIVALENT, 104 ms] (26) QDP (27) PisEmptyProof [EQUIVALENT, 0 ms] (28) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: U11(tt, V1, V2) -> U12(isNatKind(activate(V1)), activate(V1), activate(V2)) U12(tt, V1, V2) -> U13(isNatKind(activate(V2)), activate(V1), activate(V2)) U13(tt, V1, V2) -> U14(isNatKind(activate(V2)), activate(V1), activate(V2)) U14(tt, V1, V2) -> U15(isNat(activate(V1)), activate(V2)) U15(tt, V2) -> U16(isNat(activate(V2))) U16(tt) -> tt U21(tt, V1) -> U22(isNatKind(activate(V1)), activate(V1)) U22(tt, V1) -> U23(isNat(activate(V1))) U23(tt) -> tt U31(tt, V2) -> U32(isNatKind(activate(V2))) U32(tt) -> tt U41(tt) -> tt U51(tt, N) -> U52(isNatKind(activate(N)), activate(N)) U52(tt, N) -> activate(N) U61(tt, M, N) -> U62(isNatKind(activate(M)), activate(M), activate(N)) U62(tt, M, N) -> U63(isNat(activate(N)), activate(M), activate(N)) U63(tt, M, N) -> U64(isNatKind(activate(N)), activate(M), activate(N)) U64(tt, M, N) -> s(plus(activate(N), activate(M))) isNat(n__0) -> tt isNat(n__plus(V1, V2)) -> U11(isNatKind(activate(V1)), activate(V1), activate(V2)) isNat(n__s(V1)) -> U21(isNatKind(activate(V1)), activate(V1)) isNatKind(n__0) -> tt isNatKind(n__plus(V1, V2)) -> U31(isNatKind(activate(V1)), activate(V2)) isNatKind(n__s(V1)) -> U41(isNatKind(activate(V1))) plus(N, 0) -> U51(isNat(N), N) plus(N, s(M)) -> U61(isNat(M), M, N) 0 -> n__0 plus(X1, X2) -> n__plus(X1, X2) s(X) -> n__s(X) activate(n__0) -> 0 activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2)) activate(n__s(X)) -> s(activate(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: U11^1(tt, V1, V2) -> U12^1(isNatKind(activate(V1)), activate(V1), activate(V2)) U11^1(tt, V1, V2) -> ISNATKIND(activate(V1))
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