details

property	value
status	complete
benchmark	#3.21.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n105.star.cs.uiowa.edu
space	AG01

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.0872581005096 seconds
cpu usage	0.082956049
max memory	3854336.0

stage attributes

key	value
output-size	3723
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 1 : [] --> o plus : [o * o] --> o times : [o * o] --> o times(X, plus(Y, 1)) => plus(times(X, plus(Y, times(1, 0))), X) times(X, 1) => X plus(X, 0) => X times(X, 0) => 0 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] times#(X, plus(Y, 1)) =#> plus#(times(X, plus(Y, times(1, 0))), X) 1] times#(X, plus(Y, 1)) =#> times#(X, plus(Y, times(1, 0))) 2] times#(X, plus(Y, 1)) =#> plus#(Y, times(1, 0)) 3] times#(X, plus(Y, 1)) =#> times#(1, 0) Rules R_0: times(X, plus(Y, 1)) => plus(times(X, plus(Y, times(1, 0))), X) times(X, 1) => X plus(X, 0) => X times(X, 0) => 0 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, 2, 3 * 2 : * 3 : This graph has the following strongly connected components: P_1: times#(X, plus(Y, 1)) =#> times#(X, plus(Y, times(1, 0))) 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 will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: times#(X, plus(Y, 1)) >? times#(X, plus(Y, times(1, 0))) times(X, plus(Y, 1)) >= plus(times(X, plus(Y, times(1, 0))), X) times(X, 1) >= X plus(X, 0) >= X times(X, 0) >= 0 We orient these requirements with a polynomial interpretation in the natural numbers. We consider usable_rules with respect to the following argument filtering: This leaves the following ordering requirements: times#(X, plus(Y, 1)) > times#(X, plus(Y, times(1, 0))) plus(X, 0) >= X times(X, 0) >= 0 The following interpretation satisfies the requirements: 0 = 0 1 = 2 plus = \y0y1.y0 + 2y1 times = \y0y1.0 times# = \y0y1.y1 Using this interpretation, the requirements translate to: [[times#(_x0, plus(_x1, 1))]] = 4 + x1 > x1 = [[times#(_x0, plus(_x1, times(1, 0)))]] [[plus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[times(_x0, 0)]] = 0 >= 0 = [[0]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_0) by ({}, R_0). 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. popout

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