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TRS Stand 20472 pair #381710063
details
property
value
status
complete
benchmark
IJCAR_18.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n086.star.cs.uiowa.edu
space
AProVE_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.237378120422 seconds
cpu usage
0.231484216
max memory
1.01376E7
stage attributes
key
value
output-size
10698
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o div : [o * o] --> o divides : [o * o] --> o eq : [o * o] --> o false : [] --> o if : [o * o * o] --> o plus : [o * o] --> o pr : [o * o] --> o prime : [o] --> o quot : [o * o * o] --> o s : [o] --> o times : [o * o] --> o true : [] --> o plus(X, 0) => X plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) times(0, X) => 0 times(s(0), X) => X times(s(X), Y) => plus(Y, times(X, Y)) div(0, X) => 0 div(X, Y) => quot(X, Y, Y) quot(0, s(X), Y) => 0 quot(s(X), s(Y), Z) => quot(X, Y, Z) quot(X, 0, s(Y)) => s(div(X, s(Y))) div(div(X, Y), Z) => div(X, times(Y, Z)) eq(0, 0) => true eq(s(X), 0) => false eq(0, s(X)) => false eq(s(X), s(Y)) => eq(X, Y) divides(X, Y) => eq(Y, times(div(Y, X), X)) prime(s(s(X))) => pr(s(s(X)), s(X)) pr(X, s(0)) => true pr(X, s(s(Y))) => if(divides(s(s(Y)), X), X, s(Y)) if(true, X, Y) => false if(false, X, Y) => pr(X, Y) We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] plus#(s(X), Y) =#> plus#(X, Y) 1] times#(s(X), Y) =#> plus#(Y, times(X, Y)) 2] times#(s(X), Y) =#> times#(X, Y) 3] div#(X, Y) =#> quot#(X, Y, Y) 4] quot#(s(X), s(Y), Z) =#> quot#(X, Y, Z) 5] quot#(X, 0, s(Y)) =#> div#(X, s(Y)) 6] div#(div(X, Y), Z) =#> div#(X, times(Y, Z)) 7] div#(div(X, Y), Z) =#> times#(Y, Z) 8] eq#(s(X), s(Y)) =#> eq#(X, Y) 9] divides#(X, Y) =#> eq#(Y, times(div(Y, X), X)) 10] divides#(X, Y) =#> times#(div(Y, X), X) 11] divides#(X, Y) =#> div#(Y, X) 12] prime#(s(s(X))) =#> pr#(s(s(X)), s(X)) 13] pr#(X, s(s(Y))) =#> if#(divides(s(s(Y)), X), X, s(Y)) 14] pr#(X, s(s(Y))) =#> divides#(s(s(Y)), X) 15] if#(false, X, Y) =#> pr#(X, Y) Rules R_0: plus(X, 0) => X plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) times(0, X) => 0 times(s(0), X) => X times(s(X), Y) => plus(Y, times(X, Y)) div(0, X) => 0 div(X, Y) => quot(X, Y, Y) quot(0, s(X), Y) => 0 quot(s(X), s(Y), Z) => quot(X, Y, Z) quot(X, 0, s(Y)) => s(div(X, s(Y))) div(div(X, Y), Z) => div(X, times(Y, Z)) eq(0, 0) => true eq(s(X), 0) => false eq(0, s(X)) => false eq(s(X), s(Y)) => eq(X, Y) divides(X, Y) => eq(Y, times(div(Y, X), X)) prime(s(s(X))) => pr(s(s(X)), s(X)) pr(X, s(0)) => true pr(X, s(s(Y))) => if(divides(s(s(Y)), X), X, s(Y)) if(true, X, Y) => false if(false, X, Y) => pr(X, Y) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative).
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