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HRS union beta 16688 pair #381734312
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
n004.star.cs.uiowa.edu
space
Mixed_HO_12
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.0694050788879 seconds
cpu usage
0.055321166
max memory
2719744.0
stage attributes
key
value
output-size
3366
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- 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.
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