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TRS Stand 20472 pair #381716585
details
property
value
status
complete
benchmark
Ex49_GM04_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n038.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.362163066864 seconds
cpu usage
0.348493083
max memory
1.2570624E7
stage attributes
key
value
output-size
16221
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 a!6220!6220div : [o * o] --> o a!6220!6220geq : [o * o] --> o a!6220!6220if : [o * o * o] --> o a!6220!6220minus : [o * o] --> o div : [o * o] --> o false : [] --> o geq : [o * o] --> o if : [o * o * o] --> o mark : [o] --> o minus : [o * o] --> o s : [o] --> o true : [] --> o a!6220!6220minus(0, X) => 0 a!6220!6220minus(s(X), s(Y)) => a!6220!6220minus(X, Y) a!6220!6220geq(X, 0) => true a!6220!6220geq(0, s(X)) => false a!6220!6220geq(s(X), s(Y)) => a!6220!6220geq(X, Y) a!6220!6220div(0, s(X)) => 0 a!6220!6220div(s(X), s(Y)) => a!6220!6220if(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0) a!6220!6220if(true, X, Y) => mark(X) a!6220!6220if(false, X, Y) => mark(Y) mark(minus(X, Y)) => a!6220!6220minus(X, Y) mark(geq(X, Y)) => a!6220!6220geq(X, Y) mark(div(X, Y)) => a!6220!6220div(mark(X), Y) mark(if(X, Y, Z)) => a!6220!6220if(mark(X), Y, Z) mark(0) => 0 mark(s(X)) => s(mark(X)) mark(true) => true mark(false) => false a!6220!6220minus(X, Y) => minus(X, Y) a!6220!6220geq(X, Y) => geq(X, Y) a!6220!6220div(X, Y) => div(X, Y) a!6220!6220if(X, Y, Z) => if(X, Y, Z) 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] a!6220!6220minus#(s(X), s(Y)) =#> a!6220!6220minus#(X, Y) 1] a!6220!6220geq#(s(X), s(Y)) =#> a!6220!6220geq#(X, Y) 2] a!6220!6220div#(s(X), s(Y)) =#> a!6220!6220if#(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0) 3] a!6220!6220div#(s(X), s(Y)) =#> a!6220!6220geq#(X, Y) 4] a!6220!6220if#(true, X, Y) =#> mark#(X) 5] a!6220!6220if#(false, X, Y) =#> mark#(Y) 6] mark#(minus(X, Y)) =#> a!6220!6220minus#(X, Y) 7] mark#(geq(X, Y)) =#> a!6220!6220geq#(X, Y) 8] mark#(div(X, Y)) =#> a!6220!6220div#(mark(X), Y) 9] mark#(div(X, Y)) =#> mark#(X) 10] mark#(if(X, Y, Z)) =#> a!6220!6220if#(mark(X), Y, Z) 11] mark#(if(X, Y, Z)) =#> mark#(X) 12] mark#(s(X)) =#> mark#(X) Rules R_0: a!6220!6220minus(0, X) => 0 a!6220!6220minus(s(X), s(Y)) => a!6220!6220minus(X, Y) a!6220!6220geq(X, 0) => true a!6220!6220geq(0, s(X)) => false a!6220!6220geq(s(X), s(Y)) => a!6220!6220geq(X, Y) a!6220!6220div(0, s(X)) => 0 a!6220!6220div(s(X), s(Y)) => a!6220!6220if(a!6220!6220geq(X, Y), s(div(minus(X, Y), s(Y))), 0) a!6220!6220if(true, X, Y) => mark(X) a!6220!6220if(false, X, Y) => mark(Y) mark(minus(X, Y)) => a!6220!6220minus(X, Y) mark(geq(X, Y)) => a!6220!6220geq(X, Y) mark(div(X, Y)) => a!6220!6220div(mark(X), Y) mark(if(X, Y, Z)) => a!6220!6220if(mark(X), Y, Z) mark(0) => 0 mark(s(X)) => s(mark(X)) mark(true) => true mark(false) => false a!6220!6220minus(X, Y) => minus(X, Y) a!6220!6220geq(X, Y) => geq(X, Y) a!6220!6220div(X, Y) => div(X, Y) a!6220!6220if(X, Y, Z) => if(X, Y, Z) 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 : 0 * 1 : 1 * 2 : 4, 5
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