Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z118)

The rewrite relation of the following TRS is considered.

a(x1)	→	g(d(x1))	(1)
b(b(b(x1)))	→	c(d(c(x1)))	(2)
b(b(x1))	→	a(g(g(x1)))	(3)
c(d(x1))	→	g(g(x1))	(4)
g(g(g(x1)))	→	b(b(x1))	(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.

b^#(b(b(x1)))	→	c^#(x1)	(6)
b^#(b(x1))	→	g^#(g(x1))	(7)
c^#(d(x1))	→	g^#(g(x1))	(8)
b^#(b(x1))	→	g^#(x1)	(9)
c^#(d(x1))	→	g^#(x1)	(10)
b^#(b(x1))	→	a^#(g(g(x1)))	(11)
g^#(g(g(x1)))	→	b^#(b(x1))	(12)
b^#(b(b(x1)))	→	c^#(d(c(x1)))	(13)
a^#(x1)	→	g^#(d(x1))	(14)
g^#(g(g(x1)))	→	b^#(x1)	(15)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

g^#(g(g(x1)))	→	b^#(x1)	(15)
b^#(b(b(x1)))	→	c^#(d(c(x1)))	(13)
g^#(g(g(x1)))	→	b^#(b(x1))	(12)
c^#(d(x1))	→	g^#(g(x1))	(8)
b^#(b(x1))	→	g^#(g(x1))	(7)
c^#(d(x1))	→	g^#(x1)	(10)
b^#(b(x1))	→	g^#(x1)	(9)
b^#(b(b(x1)))	→	c^#(x1)	(6)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a(x₁)]	=	x₁ + 23600
[d(x₁)]	=	x₁ + 0
[b(x₁)]	=	x₁ + 35400
[c(x₁)]	=	x₁ + 47200
[g^#(x₁)]	=	x₁ + 0
[c^#(x₁)]	=	x₁ + 23601
[a^#(x₁)]	=	0
[g(x₁)]	=	x₁ + 23600
[b^#(x₁)]	=	x₁ + 2

together with the usable rules

c(d(x1))	→	g(g(x1))	(4)
a(x1)	→	g(d(x1))	(1)
b(b(x1))	→	a(g(g(x1)))	(3)
g(g(g(x1)))	→	b(b(x1))	(5)
b(b(b(x1)))	→	c(d(c(x1)))	(2)

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

g^#(g(g(x1)))	→	b^#(x1)	(15)
b^#(b(b(x1)))	→	c^#(d(c(x1)))	(13)
g^#(g(g(x1)))	→	b^#(b(x1))	(12)
c^#(d(x1))	→	g^#(g(x1))	(8)
b^#(b(x1))	→	g^#(g(x1))	(7)
c^#(d(x1))	→	g^#(x1)	(10)
b^#(b(x1))	→	g^#(x1)	(9)
b^#(b(b(x1)))	→	c^#(x1)	(6)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.