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TRS Stand 20472 pair #381712509
details
property
value
status
complete
benchmark
31.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n107.star.cs.uiowa.edu
space
Der95
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0671689510345 seconds
cpu usage
0.064097034
max memory
3821568.0
stage attributes
key
value
output-size
3477
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. !colon : [o * o] --> o !plus : [o * o] --> o a : [] --> o f : [o] --> o g : [o * o] --> o !colon(!colon(X, Y), Z) => !colon(X, !colon(Y, Z)) !colon(!plus(X, Y), Z) => !plus(!colon(X, Z), !colon(Y, Z)) !colon(X, !plus(Y, f(Z))) => !colon(g(X, Z), !plus(Y, a)) 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] !colon#(!colon(X, Y), Z) =#> !colon#(X, !colon(Y, Z)) 1] !colon#(!colon(X, Y), Z) =#> !colon#(Y, Z) 2] !colon#(!plus(X, Y), Z) =#> !colon#(X, Z) 3] !colon#(!plus(X, Y), Z) =#> !colon#(Y, Z) 4] !colon#(X, !plus(Y, f(Z))) =#> !colon#(g(X, Z), !plus(Y, a)) Rules R_0: !colon(!colon(X, Y), Z) => !colon(X, !colon(Y, Z)) !colon(!plus(X, Y), Z) => !plus(!colon(X, Z), !colon(Y, Z)) !colon(X, !plus(Y, f(Z))) => !colon(g(X, Z), !plus(Y, a)) 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, 2, 3, 4 * 1 : 0, 1, 2, 3, 4 * 2 : 0, 1, 2, 3, 4 * 3 : 0, 1, 2, 3, 4 * 4 : This graph has the following strongly connected components: P_1: !colon#(!colon(X, Y), Z) =#> !colon#(X, !colon(Y, Z)) !colon#(!colon(X, Y), Z) =#> !colon#(Y, Z) !colon#(!plus(X, Y), Z) =#> !colon#(X, Z) !colon#(!plus(X, Y), Z) =#> !colon#(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(!colon#) = 1 Thus, we can orient the dependency pairs as follows: nu(!colon#(!colon(X, Y), Z)) = !colon(X, Y) |> X = nu(!colon#(X, !colon(Y, Z))) nu(!colon#(!colon(X, Y), Z)) = !colon(X, Y) |> Y = nu(!colon#(Y, Z)) nu(!colon#(!plus(X, Y), Z)) = !plus(X, Y) |> X = nu(!colon#(X, Z)) nu(!colon#(!plus(X, Y), Z)) = !plus(X, Y) |> Y = nu(!colon#(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. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. [KusIsoSakBla09] K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui. Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems. In volume 92(10) of IEICE Transactions on Information and Systems. 2007--2015, 2009.
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