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. popout

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