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TRS Stand 20472 pair #381715180
details
property
value
status
complete
benchmark
Ex49_GM04_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n016.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.166231870651 seconds
cpu usage
0.16244387
max memory
7176192.0
stage attributes
key
value
output-size
8311
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o activate : [o] --> o div : [o * o] --> o false : [] --> o geq : [o * o] --> o if : [o * o * o] --> o minus : [o * o] --> o n!6220!62200 : [] --> o n!6220!6220s : [o] --> o s : [o] --> o true : [] --> o minus(n!6220!62200, X) => 0 minus(n!6220!6220s(X), n!6220!6220s(Y)) => minus(activate(X), activate(Y)) geq(X, n!6220!62200) => true geq(n!6220!62200, n!6220!6220s(X)) => false geq(n!6220!6220s(X), n!6220!6220s(Y)) => geq(activate(X), activate(Y)) div(0, n!6220!6220s(X)) => 0 div(s(X), n!6220!6220s(Y)) => if(geq(X, activate(Y)), n!6220!6220s(div(minus(X, activate(Y)), n!6220!6220s(activate(Y)))), n!6220!62200) if(true, X, Y) => activate(X) if(false, X, Y) => activate(Y) 0 => n!6220!62200 s(X) => n!6220!6220s(X) activate(n!6220!62200) => 0 activate(n!6220!6220s(X)) => s(X) activate(X) => X 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] minus#(n!6220!62200, X) =#> 0# 1] minus#(n!6220!6220s(X), n!6220!6220s(Y)) =#> minus#(activate(X), activate(Y)) 2] minus#(n!6220!6220s(X), n!6220!6220s(Y)) =#> activate#(X) 3] minus#(n!6220!6220s(X), n!6220!6220s(Y)) =#> activate#(Y) 4] geq#(n!6220!6220s(X), n!6220!6220s(Y)) =#> geq#(activate(X), activate(Y)) 5] geq#(n!6220!6220s(X), n!6220!6220s(Y)) =#> activate#(X) 6] geq#(n!6220!6220s(X), n!6220!6220s(Y)) =#> activate#(Y) 7] div#(0, n!6220!6220s(X)) =#> 0# 8] div#(s(X), n!6220!6220s(Y)) =#> if#(geq(X, activate(Y)), n!6220!6220s(div(minus(X, activate(Y)), n!6220!6220s(activate(Y)))), n!6220!62200) 9] div#(s(X), n!6220!6220s(Y)) =#> geq#(X, activate(Y)) 10] div#(s(X), n!6220!6220s(Y)) =#> activate#(Y) 11] div#(s(X), n!6220!6220s(Y)) =#> div#(minus(X, activate(Y)), n!6220!6220s(activate(Y))) 12] div#(s(X), n!6220!6220s(Y)) =#> minus#(X, activate(Y)) 13] div#(s(X), n!6220!6220s(Y)) =#> activate#(Y) 14] div#(s(X), n!6220!6220s(Y)) =#> activate#(Y) 15] if#(true, X, Y) =#> activate#(X) 16] if#(false, X, Y) =#> activate#(Y) 17] activate#(n!6220!62200) =#> 0# 18] activate#(n!6220!6220s(X)) =#> s#(X) Rules R_0: minus(n!6220!62200, X) => 0 minus(n!6220!6220s(X), n!6220!6220s(Y)) => minus(activate(X), activate(Y)) geq(X, n!6220!62200) => true geq(n!6220!62200, n!6220!6220s(X)) => false geq(n!6220!6220s(X), n!6220!6220s(Y)) => geq(activate(X), activate(Y)) div(0, n!6220!6220s(X)) => 0 div(s(X), n!6220!6220s(Y)) => if(geq(X, activate(Y)), n!6220!6220s(div(minus(X, activate(Y)), n!6220!6220s(activate(Y)))), n!6220!62200) if(true, X, Y) => activate(X) if(false, X, Y) => activate(Y) 0 => n!6220!62200 s(X) => n!6220!6220s(X) activate(n!6220!62200) => 0 activate(n!6220!6220s(X)) => s(X) activate(X) => X 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). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : * 1 : 0, 1, 2, 3 * 2 : 17, 18 * 3 : 17, 18 * 4 : 4, 5, 6 * 5 : 17, 18 * 6 : 17, 18 * 7 : * 8 : 15, 16 * 9 : 4, 5, 6 * 10 : 17, 18 * 11 : 7 * 12 : 0, 1, 2, 3
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