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 NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

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

(4)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a]	=	1
[b(x₁, x₂)]	=	max(x₁ + 2443, x₂ + 4885, 0)
[c(x₁, x₂, x₃)]	=	max(x₁ + 4886, x₂ + 7326, 0)
[f(x₁)]	=	x₁ + 2440
[f^#(x₁)]	=	x₁ + 0
[c^#(x₁, x₂, x₃)]	=	max(0)

together with the usable rules

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

(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))

(4)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.