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TRS Inner 89993 pair #381733542
details
property
value
status
complete
benchmark
#4.33.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n109.star.cs.uiowa.edu
space
AG01_innermost
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.77853298187 seconds
cpu usage
4.065258592
max memory
2.45055488E8
stage attributes
key
value
output-size
9932
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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, 3 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QReductionProof [EQUIVALENT, 0 ms] (9) QDP (10) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (11) QDP (12) PisEmptyProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 0 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y)) sum(cons(0, x), y) -> sum(x, y) sum(nil, y) -> y weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x))) weight(cons(n, nil)) -> n The set Q consists of the following terms: sum(cons(s(x0), x1), cons(x2, x3)) sum(cons(0, x0), x1) sum(nil, x0) weight(cons(x0, cons(x1, x2))) weight(cons(x0, nil)) ---------------------------------------- (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: SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y)) SUM(cons(0, x), y) -> SUM(x, y) WEIGHT(cons(n, cons(m, x))) -> WEIGHT(sum(cons(n, cons(m, x)), cons(0, x))) WEIGHT(cons(n, cons(m, x))) -> SUM(cons(n, cons(m, x)), cons(0, x)) The TRS R consists of the following rules: sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y)) sum(cons(0, x), y) -> sum(x, y) sum(nil, y) -> y weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x))) weight(cons(n, nil)) -> n The set Q consists of the following terms: sum(cons(s(x0), x1), cons(x2, x3)) sum(cons(0, x0), x1) sum(nil, x0) weight(cons(x0, cons(x1, x2))) weight(cons(x0, nil)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node. ---------------------------------------- (4)
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