Certification Problem

Input (TPDB TRS_Standard/SK90/2.01)

The rewrite relation of the following TRS is considered.

i(0)	→	0	(1)
+(0,y)	→	y	(2)
+(x,0)	→	x	(3)
i(i(x))	→	x	(4)
+(i(x),x)	→	0	(5)
+(x,i(x))	→	0	(6)
i(+(x,y))	→	+(i(x),i(y))	(7)
+(x,+(y,z))	→	+(+(x,y),z)	(8)
+(+(x,i(y)),y)	→	x	(9)
+(+(x,y),i(y))	→	x	(10)

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.

i^#(+(x,y))	→	i^#(x)	(11)
+^#(x,+(y,z))	→	+^#(x,y)	(12)
i^#(+(x,y))	→	i^#(y)	(13)
i^#(+(x,y))	→	+^#(i(x),i(y))	(14)
+^#(x,+(y,z))	→	+^#(+(x,y),z)	(15)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

i^#(+(x,y)) → i^#(y) (13)

i^#(+(x,y)) → i^#(x) (11)

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

[0] = 0

[i(x₁)] = 0

[+(x₁, x₂)] = x₁ + x₂ + 1

[+^#(x₁, x₂)] = 0

[i^#(x₁)] = x₁ + 0

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

i^#(+(x,y)) → i^#(y) (13)

i^#(+(x,y)) → i^#(x) (11)

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) (15)

+^#(x,+(y,z)) → +^#(x,y) (12)

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

[0] = 3

[i(x₁)] = 1

[+(x₁, x₂)] = x₁ + x₂ + 1

[+^#(x₁, x₂)] = x₂ + 0

[i^#(x₁)] = 0

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

+^#(x,+(y,z)) → +^#(+(x,y),z) (15)

+^#(x,+(y,z)) → +^#(x,y) (12)

could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.