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TRS Innermost pair #487093475
details
property
value
status
complete
benchmark
Ex1_GM99_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n139.star.cs.uiowa.edu
space
Transformed_CSR_innermost_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
3.6695 seconds
cpu usage
9.65494
user time
9.27949
system time
0.375448
max virtual memory
3.6983328E7
max residence set size
574160.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QReductionProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) QDPOrderProof [EQUIVALENT, 21 ms] (14) QDP (15) TransformationProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPQMonotonicMRRProof [EQUIVALENT, 24 ms] (18) QDP (19) TransformationProof [SOUND, 0 ms] (20) QDP (21) DependencyGraphProof [EQUIVALENT, 0 ms] (22) QDP (23) TransformationProof [EQUIVALENT, 0 ms] (24) QDP (25) DependencyGraphProof [EQUIVALENT, 0 ms] (26) QDP (27) TransformationProof [EQUIVALENT, 0 ms] (28) QDP (29) DependencyGraphProof [EQUIVALENT, 0 ms] (30) QDP (31) TransformationProof [EQUIVALENT, 0 ms] (32) QDP (33) QDPOrderProof [EQUIVALENT, 6 ms] (34) QDP (35) DependencyGraphProof [EQUIVALENT, 0 ms] (36) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(f(a, b, X)) -> mark(f(X, X, X)) active(c) -> mark(a) active(c) -> mark(b) mark(f(X1, X2, X3)) -> active(f(X1, X2, mark(X3))) mark(a) -> active(a) mark(b) -> active(b) mark(c) -> active(c) f(mark(X1), X2, X3) -> f(X1, X2, X3) f(X1, mark(X2), X3) -> f(X1, X2, X3) f(X1, X2, mark(X3)) -> f(X1, X2, X3) f(active(X1), X2, X3) -> f(X1, X2, X3) f(X1, active(X2), X3) -> f(X1, X2, X3) f(X1, X2, active(X3)) -> f(X1, X2, X3) The set Q consists of the following terms: active(f(a, b, x0)) active(c) mark(f(x0, x1, x2)) mark(a) mark(b) mark(c) f(mark(x0), x1, x2) f(x0, mark(x1), x2) f(x0, x1, mark(x2)) f(active(x0), x1, x2) f(x0, active(x1), x2) f(x0, x1, active(x2)) ---------------------------------------- (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(f(a, b, X)) -> MARK(f(X, X, X)) ACTIVE(f(a, b, X)) -> F(X, X, X) ACTIVE(c) -> MARK(a) ACTIVE(c) -> MARK(b) MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, X2, mark(X3))) MARK(f(X1, X2, X3)) -> F(X1, X2, mark(X3))
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