Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-17)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

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

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair
f^#(b(b(x,f(y)),z)) → f^#(b(b(f(a),y),y)) (8)

1.1.1 Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the arctic semiring over the integers

[f^#(x₁)] = -6 · x₁ + 0

[f(x₁)] = 2 · x₁ + 5

[a] = 3

[b(x₁, x₂)] = 0 · x₁ + 2 · x₂ + 6

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
f^#(b(b(x,f(y)),z)) → f^#(b(b(f(a),y),y)) (8)
could be deleted.
1.1.1.1 P is empty

There are no pairs anymore.