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TRS Stand 20472 pair #381712635
details
property
value
status
complete
benchmark
24.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n097.star.cs.uiowa.edu
space
Various_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0919559001923 seconds
cpu usage
0.089225352
max memory
4161536.0
stage attributes
key
value
output-size
3702
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 L : [o] --> o N : [o * o] --> o max : [o] --> o s : [o] --> o max(L(X)) => X max(N(L(0), L(X))) => X max(N(L(s(X)), L(s(Y)))) => s(max(N(L(X), L(Y)))) max(N(L(X), N(Y, Z))) => max(N(L(X), L(max(N(Y, Z))))) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): max(L(X)) >? X max(N(L(0), L(X))) >? X max(N(L(s(X)), L(s(Y)))) >? s(max(N(L(X), L(Y)))) max(N(L(X), N(Y, Z))) >? max(N(L(X), L(max(N(Y, Z))))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 L = \y0.y0 N = \y0y1.y0 + y1 max = \y0.y0 s = \y0.1 + y0 Using this interpretation, the requirements translate to: [[max(L(_x0))]] = x0 >= x0 = [[_x0]] [[max(N(L(0), L(_x0)))]] = 3 + x0 > x0 = [[_x0]] [[max(N(L(s(_x0)), L(s(_x1))))]] = 2 + x0 + x1 > 1 + x0 + x1 = [[s(max(N(L(_x0), L(_x1))))]] [[max(N(L(_x0), N(_x1, _x2)))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[max(N(L(_x0), L(max(N(_x1, _x2)))))]] We can thus remove the following rules: max(N(L(0), L(X))) => X max(N(L(s(X)), L(s(Y)))) => s(max(N(L(X), L(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] max#(N(L(X), N(Y, Z))) =#> max#(N(L(X), L(max(N(Y, Z))))) 1] max#(N(L(X), N(Y, Z))) =#> max#(N(Y, Z)) Rules R_0: max(L(X)) => X max(N(L(X), N(Y, Z))) => max(N(L(X), L(max(N(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 : * 1 : 0, 1 This graph has the following strongly connected components: P_1: max#(N(L(X), N(Y, Z))) =#> max#(N(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). 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(max#) = 1 Thus, we can orient the dependency pairs as follows: nu(max#(N(L(X), N(Y, Z)))) = N(L(X), N(Y, Z)) |> N(Y, Z) = nu(max#(N(Y, Z))) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
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