Certification Problem

Input (TPDB TRS_Standard/SK90/2.04)

The rewrite relation of the following TRS is considered.

f(+(x,0))	→	f(x)	(1)
+(x,+(y,z))	→	+(+(x,y),z)	(2)

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.

+^#(x,+(y,z))	→	+^#(x,y)	(3)
f^#(+(x,0))	→	f^#(x)	(4)
+^#(x,+(y,z))	→	+^#(+(x,y),z)	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
f^#(+(x,0)) → f^#(x) (4)

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

[f(x₁)] = 0

[0] = 1

[f^#(x₁)] = x₁ + 0

[+(x₁, x₂)] = x₁ + x₂ + 0

[+^#(x₁, x₂)] = 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
f^#(+(x,0)) → f^#(x) (4)
could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.
The 2^nd component contains the pair

+^#(x,+(y,z)) → +^#(+(x,y),z) (5)

+^#(x,+(y,z)) → +^#(x,y) (3)

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

[f(x₁)] = 0

[0] = 1

[f^#(x₁)] = 0

[+(x₁, x₂)] = x₁ + x₂ + 1

[+^#(x₁, x₂)] = x₂ + 0

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

+^#(x,+(y,z)) → +^#(+(x,y),z) (5)

+^#(x,+(y,z)) → +^#(x,y) (3)

could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.