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 AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

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

1.1 Instantiation Processor

We instantiate the pair to the following set of pairs

b^#(y_3,0)	→	a^#(c(0),a(0,y_3))	(9)
b^#(0,y_0)	→	a^#(c(y_0),a(0,0))	(10)

1.1.1 Instantiation Processor

We instantiate the pair to the following set of pairs

b^#(y_3,0)	→	a^#(0,y_3)	(11)
b^#(0,y_0)	→	a^#(0,0)	(12)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

b^#(y_3,0)	→	a^#(c(0),a(0,y_3))	(9)
a^#(y,c(b(a(0,x),0)))	→	b^#(a(c(b(0,y)),x),0)	(6)
b^#(0,y_0)	→	a^#(c(y_0),a(0,0))	(10)
a^#(y,c(b(a(0,x),0)))	→	a^#(c(b(0,y)),x)	(7)
a^#(y,c(b(a(0,x),0)))	→	b^#(0,y)	(8)
b^#(y_3,0)	→	a^#(0,y_3)	(11)

1.1.1.1.1 Reduction Pair Processor

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

[b^#(x₁, x₂)]

0	1
0	0

· x₁ +

0	1
0	0

· x₂

[0]

[a^#(x₁, x₂)]

0	1
0	0

· x₁ +

0	1
0	0

· x₂

[c(x₁)]

0	0
0	1

· x₁

[a(x₁, x₂)]

1	0
0	1

· x₁ +

0	0
0	1

· x₂

[b(x₁, x₂)]

0	0
1	1

· x₁ +

0	0
0	1

· x₂

the pairs

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

could be deleted.

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.