Certification Problem

Input (TPDB TRS_Standard/AG01/#3.39)

The rewrite relation of the following TRS is considered.

minus(x,0)	→	x	(1)
minus(s(x),s(y))	→	minus(x,y)	(2)
quot(0,s(y))	→	0	(3)
quot(s(x),s(y))	→	s(quot(minus(x,y),s(y)))	(4)
plus(0,y)	→	y	(5)
plus(s(x),y)	→	s(plus(x,y))	(6)
plus(minus(x,s(0)),minus(y,s(s(z))))	→	plus(minus(y,s(s(z))),minus(x,s(0)))	(7)

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.

plus^#(s(x),y)	→	plus^#(x,y)	(8)
quot^#(s(x),s(y))	→	minus^#(x,y)	(9)
plus^#(minus(x,s(0)),minus(y,s(s(z))))	→	plus^#(minus(y,s(s(z))),minus(x,s(0)))	(10)
minus^#(s(x),s(y))	→	minus^#(x,y)	(11)
quot^#(s(x),s(y))	→	quot^#(minus(x,y),s(y))	(12)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.

The 1^st component contains the pair

quot^#(s(x),s(y))

→

quot^#(minus(x,y),s(y))

(12)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[s(x₁)]	=	x₁ + 2
[minus(x₁, x₂)]	=	x₁ + 1
[plus^#(x₁, x₂)]	=	0
[0]	=	1
[quot(x₁, x₂)]	=	0
[minus^#(x₁, x₂)]	=	0
[plus(x₁, x₂)]	=	0
[quot^#(x₁, x₂)]	=	x₁ + 0

together with the usable rules

minus(x,0)	→	x	(1)
minus(s(x),s(y))	→	minus(x,y)	(2)

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

quot^#(s(x),s(y))

→

quot^#(minus(x,y),s(y))

(12)

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

minus^#(s(x),s(y))

→

minus^#(x,y)

(11)

1.1.2 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[s(x₁)]	=	x₁ + 2
[minus(x₁, x₂)]	=	x₁ + 1
[plus^#(x₁, x₂)]	=	0
[0]	=	1
[quot(x₁, x₂)]	=	0
[minus^#(x₁, x₂)]	=	x₁ + x₂ + 0
[plus(x₁, x₂)]	=	0
[quot^#(x₁, x₂)]	=	x₁ + 0

together with the usable rules

minus(x,0)	→	x	(1)
minus(s(x),s(y))	→	minus(x,y)	(2)

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

minus^#(s(x),s(y))

→

minus^#(x,y)

(11)

could be deleted.

1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.

The 3^rd component contains the pair

plus^#(minus(x,s(0)),minus(y,s(s(z))))	→	plus^#(minus(y,s(s(z))),minus(x,s(0)))	(10)
plus^#(s(x),y)	→	plus^#(x,y)	(8)

1.1.3 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[s(x₁)]	=	x₁ + 2
[minus(x₁, x₂)]	=	x₁ + x₂ + 1
[plus^#(x₁, x₂)]	=	x₁ + x₂ + 0
[0]	=	0
[quot(x₁, x₂)]	=	0
[minus^#(x₁, x₂)]	=	0
[plus(x₁, x₂)]	=	0
[quot^#(x₁, x₂)]	=	x₁ + 0

together with the usable rules

minus(x,0)	→	x	(1)
minus(s(x),s(y))	→	minus(x,y)	(2)

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

plus^#(s(x),y)

→

plus^#(x,y)

(8)

could be deleted.

1.1.3.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

plus^#(minus(x,s(0)),minus(y,s(s(z))))

→

plus^#(minus(y,s(s(z))),minus(x,s(0)))

(10)

1.1.3.1.1 Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals

[s(x₁)]

1	0
1	1

· x₁ +

[minus(x₁, x₂)]

1	1
1	1

· x₁ +

0	1
1	0

· x₂ +

[plus^#(x₁, x₂)]

0	1
1	1

· x₁ +

1	0
1	1

· x₂ +

[0]

[quot(x₁, x₂)]

[minus^#(x₁, x₂)]

[plus(x₁, x₂)]

[quot^#(x₁, x₂)]

together with the usable rules

minus(x,0)	→	x	(1)
minus(s(x),s(y))	→	minus(x,y)	(2)

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

plus^#(minus(x,s(0)),minus(y,s(s(z))))

→

plus^#(minus(y,s(s(z))),minus(x,s(0)))

(10)

could be deleted.

1.1.3.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.