details

property	value
status	complete
benchmark	if.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n151.star.cs.uiowa.edu
space	Mixed_HO_10

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.436009 seconds
cpu usage	0.343887
user time	0.275834
system time	0.068053
max virtual memory	113188.0
max residence set size	9540.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> nat false : [] --> bool h : [nat -> bool * nat -> nat * nat] --> nat if : [bool * nat * nat] --> nat s : [nat] --> nat true : [] --> bool Rules: if(true, x, y) => x if(false, x, y) => y h(f, g, 0) => 0 h(f, g, s(x)) => g h(f, g, if(f x, x, 0)) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We observe that the rules contain a first-order subset: if(true, X, Y) => X if(false, X, Y) => Y Moreover, the system is orthogonal. Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is terminating when seen as a many-sorted first-order TRS. According to the external first-order termination prover, this system is indeed terminating: || Input TRS: || 1: if(true(),PeRCenTX,PeRCenTY) -> PeRCenTX || 2: if(false(),PeRCenTX,PeRCenTY) -> PeRCenTY || Number of strict rules: 2 || Direct POLO(bPol) ... removes: 1 2 || false w: 1 || true w: 1 || if w: 2 * x1 + 2 * x2 + 2 * x3 || Number of strict rules: 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 and accessible arguments in [FuhKop19]). We thus obtain the following dependency pair problem (P_0, R_0, computable, formative): Dependency Pairs P_0: 0] h#(F, G, s(X)) =#> h#(F, G, if(F X, X, 0)) 1] h#(F, G, s(X)) =#> if#(F X, X, 0) Rules R_0: if(true, X, Y) => X if(false, X, Y) => Y h(F, G, 0) => 0 h(F, G, s(X)) => G h(F, G, if(F X, X, 0)) Thus, the original system is terminating if (P_0, R_0, computable, formative) is finite. We consider the dependency pair problem (P_0, R_0, computable, 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 * 1 : This graph has the following strongly connected components: P_1: h#(F, G, s(X)) =#> h#(F, G, if(F X, X, 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, computable, formative) is finite. We consider the dependency pair problem (P_1, R_0, computable, formative). The formative rules of (P_1, R_0) are R_1 ::= if(true, X, Y) => X if(false, X, Y) => Y h(F, G, s(X)) => G h(F, G, if(F X, X, 0)) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, computable, formative) by (P_1, R_1, computable, formative). Thus, the original system is terminating if (P_1, R_1, computable, formative) is finite. We consider the dependency pair problem (P_1, R_1, computable, 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: h#(F, G, s(X)) >? h#(F, G, if(F X, X, 0)) if(true, X, Y) >= X if(false, X, Y) >= Y h(F, G, s(X)) >= G h(F, G, if(F X, X, 0)) Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.) We use a recursive path ordering as defined in [Kop12, Chapter 5]. popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to Higher Order Rewriting Union Beta