Certification Problem

Input (TPDB TRS_Standard/Various_04/26)

The rewrite relation of the following TRS is considered.

f(f(x))	→	f(g(f(x),x))	(1)
f(f(x))	→	f(h(f(x),f(x)))	(2)
g(x,y)	→	y	(3)
h(x,x)	→	g(x,0)	(4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

f^#(f(x))	→	f^#(g(f(x),x))	(5)
f^#(f(x))	→	g^#(f(x),x)	(6)
f^#(f(x))	→	f^#(h(f(x),f(x)))	(7)
f^#(f(x))	→	h^#(f(x),f(x))	(8)
h^#(x,x)	→	g^#(x,0)	(9)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(f(x)) → f^#(h(f(x),f(x))) (7)

f^#(f(x)) → f^#(g(f(x),x)) (5)

1.1.1 Reduction Pair Processor with Usable Rules
Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions

prec(f) = 0 weight(f) = 2

prec(h) = 2 weight(h) = 2

prec(0) = 1 weight(0) = 1

in combination with the following argument filter

π(f^#) = 1

π(f) = [1]

π(h) = []

π(g) = 2

π(0) = []

together with the usable rules

h(x,x) → g(x,0) (4)

g(x,y) → y (3)

(w.r.t. the implicit argument filter of the reduction pair), the pairs

f^#(f(x)) → f^#(h(f(x),f(x))) (7)

f^#(f(x)) → f^#(g(f(x),x)) (5)

could be deleted.
1.1.1.1 P is empty

There are no pairs anymore.