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TRS Standard pair #487071273
details
property
value
status
complete
benchmark
Ex3_3_25_Bor03_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n179.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.29279 seconds
cpu usage
5.26795
user time
5.00221
system time
0.265738
max virtual memory
1.8478552E7
max residence set size
483224.0
stage attributes
key
value
starexec-result
NO
output
NO 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) NonTerminationLoopProof [COMPLETE, 0 ms] (9) NO (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(nil, YS) -> YS app(cons(X, XS), YS) -> cons(X, n__app(activate(XS), YS)) from(X) -> cons(X, n__from(s(X))) zWadr(nil, YS) -> nil zWadr(XS, nil) -> nil zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS))) prefix(L) -> cons(nil, n__zWadr(L, prefix(L))) app(X1, X2) -> n__app(X1, X2) from(X) -> n__from(X) nil -> n__nil zWadr(X1, X2) -> n__zWadr(X1, X2) activate(n__app(X1, X2)) -> app(X1, X2) activate(n__from(X)) -> from(X) activate(n__nil) -> nil activate(n__zWadr(X1, X2)) -> zWadr(X1, X2) activate(X) -> 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: APP(cons(X, XS), YS) -> ACTIVATE(XS) ZWADR(cons(X, XS), cons(Y, YS)) -> APP(Y, cons(X, n__nil)) ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS) ZWADR(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS) PREFIX(L) -> NIL PREFIX(L) -> PREFIX(L) ACTIVATE(n__app(X1, X2)) -> APP(X1, X2) ACTIVATE(n__from(X)) -> FROM(X) ACTIVATE(n__nil) -> NIL ACTIVATE(n__zWadr(X1, X2)) -> ZWADR(X1, X2) The TRS R consists of the following rules: app(nil, YS) -> YS app(cons(X, XS), YS) -> cons(X, n__app(activate(XS), YS)) from(X) -> cons(X, n__from(s(X))) zWadr(nil, YS) -> nil zWadr(XS, nil) -> nil zWadr(cons(X, XS), cons(Y, YS)) -> cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS))) prefix(L) -> cons(nil, n__zWadr(L, prefix(L))) app(X1, X2) -> n__app(X1, X2) from(X) -> n__from(X) nil -> n__nil zWadr(X1, X2) -> n__zWadr(X1, X2) activate(n__app(X1, X2)) -> app(X1, X2) activate(n__from(X)) -> from(X) activate(n__nil) -> nil activate(n__zWadr(X1, X2)) -> zWadr(X1, X2) activate(X) -> X Q is empty. 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 3 less nodes. ---------------------------------------- (4)
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