Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-15)

The rewrite relation of the following TRS is considered.

c(z,x,a)	→	f(b(b(f(z),z),x))	(1)
b(y,b(z,a))	→	f(b(c(f(a),y,z),z))	(2)
f(c(c(z,a,a),x,a))	→	z	(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.

b^#(y,b(z,a))	→	b^#(c(f(a),y,z),z)	(4)
c^#(z,x,a)	→	b^#(b(f(z),z),x)	(5)
c^#(z,x,a)	→	b^#(f(z),z)	(6)
c^#(z,x,a)	→	f^#(b(b(f(z),z),x))	(7)
b^#(y,b(z,a))	→	c^#(f(a),y,z)	(8)
b^#(y,b(z,a))	→	f^#(b(c(f(a),y,z),z))	(9)
c^#(z,x,a)	→	f^#(z)	(10)
b^#(y,b(z,a))	→	f^#(a)	(11)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

b^#(y,b(z,a))	→	c^#(f(a),y,z)	(8)
c^#(z,x,a)	→	b^#(b(f(z),z),x)	(5)
c^#(z,x,a)	→	b^#(f(z),z)	(6)
b^#(y,b(z,a))	→	b^#(c(f(a),y,z),z)	(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]

11106

11102

[b(x₁, x₂)]

1	0
0	0

· x₁ +

1	0
1	0

· x₂ +

[c(x₁, x₂, x₃)]

0	0
1	1

· x₁ +

1	0
1	0

· x₂ +

0	0
1	0

· x₃ +

[f(x₁)]

0	1
0	1

· x₁ +

[f^#(x₁)]

[c^#(x₁, x₂, x₃)]

1	1
1	1

· x₁ +

1	0
1	0

· x₂ +

1	0
1	0

· x₃ +

[b^#(x₁, x₂)]

1	0
1	0

· x₁ +

1	0
1	0

· x₂ +

11103

11105

together with the usable rules

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

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

b^#(y,b(z,a))	→	c^#(f(a),y,z)	(8)
c^#(z,x,a)	→	b^#(b(f(z),z),x)	(5)
c^#(z,x,a)	→	b^#(f(z),z)	(6)
b^#(y,b(z,a))	→	b^#(c(f(a),y,z),z)	(4)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.