Certification Problem

Input (TPDB TRS_Standard/SK90/2.11)

The rewrite relation of the following TRS is considered.

+(0,y)	→	y	(1)
+(s(x),y)	→	s(+(x,y))	(2)
-(0,y)	→	0	(3)
-(x,0)	→	x	(4)
-(s(x),s(y))	→	-(x,y)	(5)

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.

-^#(s(x),s(y))	→	-^#(x,y)	(6)
+^#(s(x),y)	→	+^#(x,y)	(7)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
+^#(s(x),y) → +^#(x,y) (7)

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

[s(x₁)] = x₁ + 1

[0] = 0

[-(x₁, x₂)] = 0

[-^#(x₁, x₂)] = 0

[+(x₁, x₂)] = 0

[+^#(x₁, x₂)] = x₁ + 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
+^#(s(x),y) → +^#(x,y) (7)
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
-^#(s(x),s(y)) → -^#(x,y) (6)

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

[s(x₁)] = x₁ + 1

[0] = 0

[-(x₁, x₂)] = 0

[-^#(x₁, x₂)] = x₂ + 0

[+(x₁, x₂)] = 0

[+^#(x₁, x₂)] = 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
-^#(s(x),s(y)) → -^#(x,y) (6)
could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.