Certification Problem

Input (TPDB TRS_Standard/Zantema_05/z10)

The rewrite relation of the following TRS is considered.

a(lambda(x),y)	→	lambda(a(x,p(1,a(y,t))))	(1)
a(p(x,y),z)	→	p(a(x,z),a(y,z))	(2)
a(a(x,y),z)	→	a(x,a(y,z))	(3)
a(id,x)	→	x	(4)
a(1,id)	→	1	(5)
a(t,id)	→	t	(6)
a(1,p(x,y))	→	x	(7)
a(t,p(x,y))	→	y	(8)

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.

a^#(p(x,y),z)	→	a^#(y,z)	(9)
a^#(lambda(x),y)	→	a^#(y,t)	(10)
a^#(lambda(x),y)	→	a^#(x,p(1,a(y,t)))	(11)
a^#(a(x,y),z)	→	a^#(x,a(y,z))	(12)
a^#(p(x,y),z)	→	a^#(x,z)	(13)
a^#(a(x,y),z)	→	a^#(y,z)	(14)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(a(x,y),z)	→	a^#(y,z)	(14)
a^#(p(x,y),z)	→	a^#(x,z)	(13)
a^#(a(x,y),z)	→	a^#(x,a(y,z))	(12)
a^#(lambda(x),y)	→	a^#(y,t)	(10)
a^#(lambda(x),y)	→	a^#(x,p(1,a(y,t)))	(11)
a^#(p(x,y),z)	→	a^#(y,z)	(9)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a(x₁, x₂)]	=	x₁ + x₂ + 0
[1]	=	0
[t]	=	0
[lambda(x₁)]	=	x₁ + 1
[p(x₁, x₂)]	=	max(x₁ + 0, x₂ + 0, 0)
[id]	=	11293
[a^#(x₁, x₂)]	=	x₁ + x₂ + 0

together with the usable rules

a(id,x)	→	x	(4)
a(t,p(x,y))	→	y	(8)
a(lambda(x),y)	→	lambda(a(x,p(1,a(y,t))))	(1)
a(a(x,y),z)	→	a(x,a(y,z))	(3)
a(1,id)	→	1	(5)
a(1,p(x,y))	→	x	(7)
a(t,id)	→	t	(6)
a(p(x,y),z)	→	p(a(x,z),a(y,z))	(2)

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

a^#(lambda(x),y)	→	a^#(y,t)	(10)
a^#(lambda(x),y)	→	a^#(x,p(1,a(y,t)))	(11)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(a(x,y),z) → a^#(y,z) (14)

a^#(a(x,y),z) → a^#(x,a(y,z)) (12)

a^#(p(x,y),z) → a^#(y,z) (9)

a^#(p(x,y),z) → a^#(x,z) (13)

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

[a(x₁, x₂)] = x₁ + x₂ + 1

[1] = 1

[t] = 1

[lambda(x₁)] = x₁ + 18816

[p(x₁, x₂)] = x₁ + x₂ + 1

[id] = 62898

[a^#(x₁, x₂)] = x₁ + 0

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

a^#(a(x,y),z) → a^#(y,z) (14)

a^#(a(x,y),z) → a^#(x,a(y,z)) (12)

a^#(p(x,y),z) → a^#(y,z) (9)

a^#(p(x,y),z) → a^#(x,z) (13)

could be deleted.
1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.