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TRS Stand 20472 pair #381710900
details
property
value
status
complete
benchmark
IJCAR_12.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n104.star.cs.uiowa.edu
space
AProVE_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.168882846832 seconds
cpu usage
0.113507279
max memory
4788224.0
stage attributes
key
value
output-size
6411
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 plus : [o * o] --> o quot : [o * o * o] --> o s : [o] --> o times : [o * o] --> 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)) 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) 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)) 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 : 0 * 2 : 1, 2 * 3 : 4, 5 * 4 : 4, 5 * 5 : 3, 6, 7 * 6 : 3, 6, 7 * 7 : 1, 2 This graph has the following strongly connected components: P_1: plus#(s(X), Y) =#> plus#(X, Y) P_2: times#(s(X), Y) =#> times#(X, Y) P_3: div#(X, Y) =#> quot#(X, Y, Y) quot#(s(X), s(Y), Z) =#> quot#(X, Y, Z) quot#(X, 0, s(Y)) =#> div#(X, s(Y)) div#(div(X, Y), Z) =#> div#(X, times(Y, Z)) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f), (P_2, R_0, m, f) and (P_3, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_0, minimal, formative). We apply the subterm criterion with the following projection function:
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