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TRS Standard pair #487067003
details
property
value
status
complete
benchmark
#4.13.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
Strategy_removed_AG01
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.38205 seconds
cpu usage
5.85749
user time
5.59692
system time
0.260573
max virtual memory
1.8476724E7
max residence set size
492048.0
stage attributes
key
value
starexec-result
NO
output
NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) NonTerminationLoopProof [COMPLETE, 559 ms] (4) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0, 1, x) -> f(x, x, x) f(x, y, z) -> 2 0 -> 2 1 -> 2 g(x, x, y) -> y g(x, y, y) -> 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: F(0, 1, x) -> F(x, x, x) The TRS R consists of the following rules: f(0, 1, x) -> f(x, x, x) f(x, y, z) -> 2 0 -> 2 1 -> 2 g(x, x, y) -> y g(x, y, y) -> x Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: s = F(g(1, 0, 0), g(1, y, y), x) evaluates to t =F(x, x, x) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [y / 0, x / g(1, 0, 0)] -------------------------------------------------------------------------------- Rewriting sequence F(g(1, 0, 0), g(1, 0, 0), g(1, 0, 0)) -> F(g(1, 0, 0), 1, g(1, 0, 0)) with rule g(x', y, y) -> x' at position [1] and matcher [x' / 1, y / 0] F(g(1, 0, 0), 1, g(1, 0, 0)) -> F(g(1, 2, 0), 1, g(1, 0, 0)) with rule 0 -> 2 at position [0,1] and matcher [ ] F(g(1, 2, 0), 1, g(1, 0, 0)) -> F(g(2, 2, 0), 1, g(1, 0, 0)) with rule 1 -> 2 at position [0,0] and matcher [ ] F(g(2, 2, 0), 1, g(1, 0, 0)) -> F(0, 1, g(1, 0, 0)) with rule g(x', x', y) -> y at position [0] and matcher [x' / 2, y / 0] F(0, 1, g(1, 0, 0)) -> F(g(1, 0, 0), g(1, 0, 0), g(1, 0, 0)) with rule F(0, 1, x) -> F(x, x, x) Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (4) NO
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