Certification Problem

Input (TPDB TRS_Standard/SK90/4.53)

The rewrite relation of the following TRS is considered.

f(a)	→	b	(1)
f(c)	→	d	(2)
f(g(x,y))	→	g(f(x),f(y))	(3)
f(h(x,y))	→	g(h(y,f(x)),h(x,f(y)))	(4)
g(x,x)	→	h(e,x)	(5)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

f^#(g(x,y))	→	f^#(x)	(6)
f^#(h(x,y))	→	f^#(y)	(7)
f^#(h(x,y))	→	f^#(x)	(8)
f^#(g(x,y))	→	f^#(y)	(9)
f^#(g(x,y))	→	g^#(f(x),f(y))	(10)
f^#(h(x,y))	→	g^#(h(y,f(x)),h(x,f(y)))	(11)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a]	=	0
[h(x₁, x₂)]	=	x₁ + x₂ + 1
[d]	=	0
[b]	=	0
[c]	=	0
[f(x₁)]	=	0
[f^#(x₁)]	=	x₁ + 0
[g^#(x₁, x₂)]	=	0
[e]	=	0
[g(x₁, x₂)]	=	x₁ + x₂ + 1

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pairs

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

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.