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TRS Stand 20472 pair #381712392
details
property
value
status
complete
benchmark
#4.23.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n112.star.cs.uiowa.edu
space
Strategy_removed_AG01
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0989181995392 seconds
cpu usage
0.094636651
max memory
4087808.0
stage attributes
key
value
output-size
4929
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 quot : [o * o * o] --> o s : [o] --> o quot(0, s(X), s(Y)) => 0 quot(s(X), s(Y), Z) => quot(X, Y, Z) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) quot(X, 0, s(Y)) => s(quot(X, plus(Y, s(0)), 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] quot#(s(X), s(Y), Z) =#> quot#(X, Y, Z) 1] plus#(s(X), Y) =#> plus#(X, Y) 2] quot#(X, 0, s(Y)) =#> quot#(X, plus(Y, s(0)), s(Y)) 3] quot#(X, 0, s(Y)) =#> plus#(Y, s(0)) Rules R_0: quot(0, s(X), s(Y)) => 0 quot(s(X), s(Y), Z) => quot(X, Y, Z) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) quot(X, 0, s(Y)) => s(quot(X, plus(Y, s(0)), 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, 2, 3 * 1 : 1 * 2 : 0, 2, 3 * 3 : 1 This graph has the following strongly connected components: P_1: quot#(s(X), s(Y), Z) =#> quot#(X, Y, Z) quot#(X, 0, s(Y)) =#> quot#(X, plus(Y, s(0)), s(Y)) P_2: plus#(s(X), 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#) = 1 Thus, we can orient the dependency pairs as follows: nu(plus#(s(X), Y)) = s(X) |> X = 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 apply the subterm criterion with the following projection function: nu(quot#) = 1 Thus, we can orient the dependency pairs as follows: nu(quot#(s(X), s(Y), Z)) = s(X) |> X = nu(quot#(X, Y, Z)) nu(quot#(X, 0, s(Y))) = X = X = nu(quot#(X, plus(Y, s(0)), s(Y))) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by (P_3, R_0, minimal, f), where P_3 contains: quot#(X, 0, s(Y)) =#> quot#(X, plus(Y, s(0)), s(Y)) Thus, the original system is terminating if (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_0, minimal, formative).
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