details

property	value
status	complete
benchmark	twice_modif4.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n148.star.cs.uiowa.edu
space	Mixed_HO_12

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.104536 seconds
cpu usage	0.094018
user time	0.074373
system time	0.019645
max virtual memory	113188.0
max residence set size	3980.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> nat I : [nat] --> nat s : [nat] --> nat Rules: I(s(x)) => s((/\f./\y.f (f y)) (/\z.I(z)) x) I(0) => 0 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs. We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] I#(s(X)) =#> I#(x) 1] I#(s(X)) =#> I#(X) 2] I#(s(X)) =#> I#((/\x.I(x)) X) 3] I#(s(X)) =#> I#(X) Rules R_0: I(s(X)) => s((/\f./\x.f (f x)) (/\y.I(y)) X) I(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 : 0, 1, 2, 3 * 3 : 0, 1, 2, 3 This graph has the following strongly connected components: P_1: I#(s(X)) =#> I#(X) I#(s(X)) =#> I#((/\x.I(x)) X) I#(s(X)) =#> I#(X) 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). The formative rules of (P_1, R_0) are R_1 ::= I(s(X)) => s((/\f./\x.f (f x)) (/\y.I(y)) X) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative). Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_1, 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: I#(s(X)) >? I#(X) I#(s(X)) >? I#((/\x.I(x)) X) I#(s(X)) >? I#(X) I(s(X)) >= s((/\f./\x.f (f x)) (/\y.I(y)) X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: I = \y0.y0 I# = \y0.3y0 s = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[I#(s(_x0))]] = 9 + 9x0 > 3x0 = [[I#(_x0)]] [[I#(s(_x0))]] = 9 + 9x0 > 3x0 = [[I#((/\x.I(x)) _x0)]] [[I#(s(_x0))]] = 9 + 9x0 > 3x0 = [[I#(_x0)]] [[I(s(_x0))]] = 3 + 3x0 >= 3 + 3x0 = [[s((/\f./\x.f (f x)) (/\y.I(y)) _x0)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_1) by ({}, R_1). 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. popout

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all output return to Higher Order Rewriting Union Beta