Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-1)

The rewrite relation of the following TRS is considered.

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

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

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a]	=	0
[b(x₁, x₂)]	=	max(x₁ + 26576, x₂ + 26574, 0)
[c(x₁, x₂, x₃)]	=	max(x₂ + 17718, x₃ + 1, 0)
[f(x₁)]	=	x₁ + 8858
[f^#(x₁)]	=	x₁ + 14236
[b^#(x₁, x₂)]	=	max(x₁ + 14235, x₂ + 31953, 0)

together with the usable rules

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

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

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

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.