Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/4)

The rewrite relation of the following TRS is considered.

c(c(b(c(x))))	→	b(a(0,c(x)))	(1)
c(c(x))	→	b(c(b(c(x))))	(2)
a(0,x)	→	c(c(x))	(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^#(0,x)	→	c^#(c(x))	(4)
a^#(0,x)	→	c^#(x)	(5)
c^#(c(x))	→	c^#(b(c(x)))	(6)
c^#(c(b(c(x))))	→	a^#(0,c(x))	(7)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a(x₁, x₂)]	=	x₂ + 2
[b(x₁)]	=	x₁ + 0
[c(x₁)]	=	x₁ + 1
[0]	=	1
[c^#(x₁)]	=	x₁ + 0
[a^#(x₁, x₂)]	=	x₁ + x₂ + 0

together with the usable rules

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

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

a^#(0,x)

→

c^#(x)

(5)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

c^#(c(b(c(x))))	→	a^#(0,c(x))	(7)
a^#(0,x)	→	c^#(c(x))	(4)

1.1.1.1.1 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	0

· x₁ +

1	0
1	1

· x₂ +

[b(x₁)]

1	0
0	0

· x₁ +

[c(x₁)]

0	1
1	0

· x₁ +

[0]

[c^#(x₁)]

0	1
0	0

· x₁ +

[a^#(x₁, x₂)]

0	0
1	0

· x₁ +

1	0
0	0

· x₂ +

together with the usable rules

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

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

c^#(c(b(c(x))))	→	a^#(0,c(x))	(7)
a^#(0,x)	→	c^#(c(x))	(4)

could be deleted.

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.