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TRS Stand 20472 pair #381715567
details
property
value
status
complete
benchmark
Ex6_Luc98_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n036.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.53049588203 seconds
cpu usage
6.60803246
max memory
4.50265088E8
stage attributes
key
value
output-size
27361
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 31 ms] (2) QDP (3) DependencyGraphProof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, 7 ms] (27) QDP (28) QDPOrderProof [EQUIVALENT, 9 ms] (29) QDP (30) QDPOrderProof [EQUIVALENT, 0 ms] (31) QDP (32) DependencyGraphProof [EQUIVALENT, 0 ms] (33) QDP (34) UsableRulesProof [EQUIVALENT, 0 ms] (35) QDP (36) QDPSizeChangeProof [EQUIVALENT, 0 ms] (37) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(first(0, X)) -> mark(nil) active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z))) active(from(X)) -> mark(cons(X, from(s(X)))) mark(first(X1, X2)) -> active(first(mark(X1), mark(X2))) mark(0) -> active(0) mark(nil) -> active(nil) mark(s(X)) -> active(s(mark(X))) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) first(mark(X1), X2) -> first(X1, X2) first(X1, mark(X2)) -> first(X1, X2) first(active(X1), X2) -> first(X1, X2) first(X1, active(X2)) -> first(X1, X2) s(mark(X)) -> s(X) s(active(X)) -> s(X) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(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: ACTIVE(first(0, X)) -> MARK(nil) ACTIVE(first(s(X), cons(Y, Z))) -> MARK(cons(Y, first(X, Z))) ACTIVE(first(s(X), cons(Y, Z))) -> CONS(Y, first(X, Z)) ACTIVE(first(s(X), cons(Y, Z))) -> FIRST(X, Z) ACTIVE(from(X)) -> MARK(cons(X, from(s(X))))
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