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TRS Stand 20472 pair #381712136
details
property
value
status
complete
benchmark
t014.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n036.star.cs.uiowa.edu
space
HirokawaMiddeldorp_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.107751131058 seconds
cpu usage
0.097768002
max memory
4530176.0
stage attributes
key
value
output-size
5561
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. !minus : [o * o] --> o 0 : [] --> o div : [o * o] --> o false : [] --> o if : [o * o * o] --> o lt : [o * o] --> o s : [o] --> o true : [] --> o !minus(X, 0) => X !minus(0, s(X)) => 0 !minus(s(X), s(Y)) => !minus(X, Y) lt(X, 0) => false lt(0, s(X)) => true lt(s(X), s(Y)) => lt(X, Y) if(true, X, Y) => X if(false, X, Y) => Y div(X, 0) => 0 div(0, X) => 0 div(s(X), s(Y)) => if(lt(X, Y), 0, s(div(!minus(X, Y), s(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] !minus#(s(X), s(Y)) =#> !minus#(X, Y) 1] lt#(s(X), s(Y)) =#> lt#(X, Y) 2] div#(s(X), s(Y)) =#> if#(lt(X, Y), 0, s(div(!minus(X, Y), s(Y)))) 3] div#(s(X), s(Y)) =#> lt#(X, Y) 4] div#(s(X), s(Y)) =#> div#(!minus(X, Y), s(Y)) 5] div#(s(X), s(Y)) =#> !minus#(X, Y) Rules R_0: !minus(X, 0) => X !minus(0, s(X)) => 0 !minus(s(X), s(Y)) => !minus(X, Y) lt(X, 0) => false lt(0, s(X)) => true lt(s(X), s(Y)) => lt(X, Y) if(true, X, Y) => X if(false, X, Y) => Y div(X, 0) => 0 div(0, X) => 0 div(s(X), s(Y)) => if(lt(X, Y), 0, s(div(!minus(X, Y), s(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). 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 : * 3 : 1 * 4 : 2, 3, 4, 5 * 5 : 0 This graph has the following strongly connected components: P_1: !minus#(s(X), s(Y)) =#> !minus#(X, Y) P_2: lt#(s(X), s(Y)) =#> lt#(X, Y) P_3: div#(s(X), s(Y)) =#> div#(!minus(X, Y), s(Y)) 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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_3, R_0) are: !minus(X, 0) => X !minus(0, s(X)) => 0 !minus(s(X), s(Y)) => !minus(X, Y) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient:
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