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TRS Innermost pair #487092967
details
property
value
status
complete
benchmark
#4.23.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n150.star.cs.uiowa.edu
space
AG01_innermost
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.92513 seconds
cpu usage
4.21041
user time
4.0204
system time
0.19001
max virtual memory
1.8477524E7
max residence set size
263580.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 19 ms] (2) QDP (3) DependencyGraphProof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) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QReductionProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPOrderProof [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: quot(0, s(y), s(z)) -> 0 quot(s(x), s(y), z) -> quot(x, y, z) plus(0, y) -> y plus(s(x), y) -> s(plus(x, y)) quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z))) The set Q consists of the following terms: quot(0, s(x0), s(x1)) quot(s(x0), s(x1), x2) plus(0, x0) plus(s(x0), x1) quot(x0, 0, s(x1)) ---------------------------------------- (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: QUOT(s(x), s(y), z) -> QUOT(x, y, z) PLUS(s(x), y) -> PLUS(x, y) QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z)) QUOT(x, 0, s(z)) -> PLUS(z, s(0)) The TRS R consists of the following rules: quot(0, s(y), s(z)) -> 0 quot(s(x), s(y), z) -> quot(x, y, z) plus(0, y) -> y plus(s(x), y) -> s(plus(x, y)) quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z))) The set Q consists of the following terms: quot(0, s(x0), s(x1)) quot(s(x0), s(x1), x2) plus(0, x0) plus(s(x0), x1) quot(x0, 0, s(x1)) 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) Complex Obligation (AND) ---------------------------------------- (5) Obligation:
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