Certification Problem

Input (TPDB TRS_Standard/Der95/28)

The rewrite relation of the following TRS is considered.

if(if(x,y,z),u,v)

→

if(x,if(y,u,v),if(z,u,v))

(1)

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.

if^#(if(x,y,z),u,v)	→	if^#(y,u,v)	(2)
if^#(if(x,y,z),u,v)	→	if^#(z,u,v)	(3)
if^#(if(x,y,z),u,v)	→	if^#(x,if(y,u,v),if(z,u,v))	(4)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

if^#(if(x,y,z),u,v) → if^#(x,if(y,u,v),if(z,u,v)) (4)

if^#(if(x,y,z),u,v) → if^#(z,u,v) (3)

if^#(if(x,y,z),u,v) → if^#(y,u,v) (2)

1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation

[if(x₁, x₂, x₃)] = x₁ + x₂ + x₃ + 1

[if^#(x₁, x₂, x₃)] = x₁ + 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pairs

if^#(if(x,y,z),u,v) → if^#(x,if(y,u,v),if(z,u,v)) (4)

if^#(if(x,y,z),u,v) → if^#(z,u,v) (3)

if^#(if(x,y,z),u,v) → if^#(y,u,v) (2)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.