Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/6)

The rewrite relation of the following TRS is considered.

b(x,y)	→	c(a(c(y),a(0,x)))	(1)
a(y,x)	→	y	(2)
a(y,c(b(a(0,x),0)))	→	b(a(c(b(0,y)),x),0)	(3)

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.

a^#(y,c(b(a(0,x),0)))	→	a^#(c(b(0,y)),x)	(4)
b^#(x,y)	→	a^#(0,x)	(5)
a^#(y,c(b(a(0,x),0)))	→	b^#(0,y)	(6)
a^#(y,c(b(a(0,x),0)))	→	b^#(a(c(b(0,y)),x),0)	(7)
b^#(x,y)	→	a^#(c(y),a(0,x))	(8)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

b^#(x,y)	→	a^#(c(y),a(0,x))	(8)
a^#(y,c(b(a(0,x),0)))	→	b^#(a(c(b(0,y)),x),0)	(7)
a^#(y,c(b(a(0,x),0)))	→	b^#(0,y)	(6)
b^#(x,y)	→	a^#(0,x)	(5)
a^#(y,c(b(a(0,x),0)))	→	a^#(c(b(0,y)),x)	(4)

1.1.1 Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals

[a(x₁, x₂)]

1	1
0	1

· x₁ +

1	1
1	0

· x₂ +

[b(x₁, x₂)]

1	1
1	1

· x₁ +

0	0
0	1

· x₂ +

86123

[c(x₁)]

0	1
0	0

· x₁ +

[0]

43057

43062

[a^#(x₁, x₂)]

0	1
0	0

· x₁ +

1	0
0	0

· x₂ +

[b^#(x₁, x₂)]

1	1
0	0

· x₁ +

86121

together with the usable rules

b(x,y)	→	c(a(c(y),a(0,x)))	(1)
a(y,c(b(a(0,x),0)))	→	b(a(c(b(0,y)),x),0)	(3)
a(y,x)	→	y	(2)

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

b^#(x,y)	→	a^#(c(y),a(0,x))	(8)
a^#(y,c(b(a(0,x),0)))	→	b^#(a(c(b(0,y)),x),0)	(7)
a^#(y,c(b(a(0,x),0)))	→	b^#(0,y)	(6)
b^#(x,y)	→	a^#(0,x)	(5)
a^#(y,c(b(a(0,x),0)))	→	a^#(c(b(0,y)),x)	(4)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.