details

property	value
status	complete
benchmark	#4.32.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)	0.0694601535797 seconds
cpu usage	0.05717439
max memory	2727936.0

stage attributes

key	value
output-size	3226
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. c : [o] --> o f : [o * o] --> o s : [o] --> o f(X, c(Y)) => f(X, s(f(Y, Y))) f(s(X), s(Y)) => f(X, s(c(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] f#(X, c(Y)) =#> f#(X, s(f(Y, Y))) 1] f#(X, c(Y)) =#> f#(Y, Y) 2] f#(s(X), s(Y)) =#> f#(X, s(c(s(Y)))) Rules R_0: f(X, c(Y)) => f(X, s(f(Y, Y))) f(s(X), s(Y)) => f(X, s(c(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 : 2 * 1 : 0, 1, 2 * 2 : 2 This graph has the following strongly connected components: P_1: f#(X, c(Y)) =#> f#(Y, Y) P_2: f#(s(X), s(Y)) =#> f#(X, s(c(s(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(f#) = 1 Thus, we can orient the dependency pairs as follows: nu(f#(s(X), s(Y))) = s(X) |> X = nu(f#(X, s(c(s(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(f#) = 2 Thus, we can orient the dependency pairs as follows: nu(f#(X, c(Y))) = c(Y) |> Y = nu(f#(Y, Y)) 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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