Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-14)

The rewrite relation of the following TRS is considered.

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

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

b^#(f(b(x,z)),y)	→	b^#(y,z)	(5)
b^#(f(b(x,z)),y)	→	b^#(z,b(y,z))	(4)

1.1.1 Reduction Pair Processor with Usable Rules

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

[f(x₁)]

0	0
1	0

· x₁

[b(x₁, x₂)]

1	0
0	1

· x₁ +

1	1
0	0

· x₂

[c(x₁, x₂, x₃)]

1	1
1	1

· x₁ +

1	0
1	0

· x₂ +

1	0
0	1

· x₃

[a]

together with the usable rules

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

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

b^#(f(b(x,z)),y)

→

b^#(y,z)

(5)

could be deleted.

1.1.1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

b^#(f(b(x,z)),y)	→	b^#(z,b(y,z))	(4)

1	>	1

As there is no critical graph in the transitive closure, there are no infinite chains.