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TRS Stand 20472 pair #381711853
details
property
value
status
complete
benchmark
#3.22.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n004.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.163187026978 seconds
cpu usage
0.145719809
max memory
5836800.0
stage attributes
key
value
output-size
7367
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 plus : [o * o] --> o s : [o] --> o times : [o * o] --> o times(X, plus(Y, s(Z))) => plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) times(X, 0) => 0 times(X, s(Y)) => plus(times(X, Y), X) plus(X, 0) => X plus(X, s(Y)) => s(plus(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] times#(X, plus(Y, s(Z))) =#> plus#(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) 1] times#(X, plus(Y, s(Z))) =#> times#(X, plus(Y, times(s(Z), 0))) 2] times#(X, plus(Y, s(Z))) =#> plus#(Y, times(s(Z), 0)) 3] times#(X, plus(Y, s(Z))) =#> times#(s(Z), 0) 4] times#(X, plus(Y, s(Z))) =#> times#(X, s(Z)) 5] times#(X, s(Y)) =#> plus#(times(X, Y), X) 6] times#(X, s(Y)) =#> times#(X, Y) 7] plus#(X, s(Y)) =#> plus#(X, Y) Rules R_0: times(X, plus(Y, s(Z))) => plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) times(X, 0) => 0 times(X, s(Y)) => plus(times(X, Y), X) plus(X, 0) => X plus(X, s(Y)) => s(plus(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). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 7 * 1 : 0, 1, 2, 3, 4, 5, 6 * 2 : 7 * 3 : * 4 : 5, 6 * 5 : 7 * 6 : 0, 1, 2, 3, 4, 5, 6 * 7 : 7 This graph has the following strongly connected components: P_1: times#(X, plus(Y, s(Z))) =#> times#(X, plus(Y, times(s(Z), 0))) times#(X, plus(Y, s(Z))) =#> times#(X, s(Z)) times#(X, s(Y)) =#> times#(X, Y) P_2: plus#(X, s(Y)) =#> plus#(X, 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) and (P_2, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(plus#) = 2 Thus, we can orient the dependency pairs as follows: nu(plus#(X, s(Y))) = s(Y) |> Y = nu(plus#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: times#(X, plus(Y, s(Z))) >? times#(X, plus(Y, times(s(Z), 0))) times#(X, plus(Y, s(Z))) >? times#(X, s(Z)) times#(X, s(Y)) >? times#(X, Y) times(X, plus(Y, s(Z))) >= plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) times(X, 0) >= 0 times(X, s(Y)) >= plus(times(X, Y), X)
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