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TRS Inner 89993 pair #381733993
details
property
value
status
complete
benchmark
Ex1_2_Luc02c_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n071.star.cs.uiowa.edu
space
Transformed_CSR_innermost_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.7332470417 seconds
cpu usage
7.402677921
max memory
4.355072E8
stage attributes
key
value
output-size
11319
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 10 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 63 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 31 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 16 ms] (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QReductionProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__2nd(cons(X, cons(Y, Z))) -> mark(Y) a__from(X) -> cons(mark(X), from(s(X))) mark(2nd(X)) -> a__2nd(mark(X)) mark(from(X)) -> a__from(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) a__2nd(X) -> 2nd(X) a__from(X) -> from(X) The set Q consists of the following terms: a__from(x0) mark(2nd(x0)) mark(from(x0)) mark(cons(x0, x1)) mark(s(x0)) a__2nd(x0) ---------------------------------------- (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: A__2ND(cons(X, cons(Y, Z))) -> MARK(Y) A__FROM(X) -> MARK(X) MARK(2nd(X)) -> A__2ND(mark(X)) MARK(2nd(X)) -> MARK(X) MARK(from(X)) -> A__FROM(mark(X)) MARK(from(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__2nd(cons(X, cons(Y, Z))) -> mark(Y) a__from(X) -> cons(mark(X), from(s(X))) mark(2nd(X)) -> a__2nd(mark(X)) mark(from(X)) -> a__from(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) a__2nd(X) -> 2nd(X) a__from(X) -> from(X) The set Q consists of the following terms: a__from(x0) mark(2nd(x0)) mark(from(x0)) mark(cons(x0, x1)) mark(s(x0)) a__2nd(x0) We have to consider all minimal (P,Q,R)-chains. ----------------------------------------
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