Certification Problem

Input (TPDB TRS_Standard/Endrullis_06/linear2)

The rewrite relation of the following TRS is considered.

f(a,f(a,f(a,f(x,b))))	→	f(f(a,f(a,f(a,x))),b)	(1)
f(f(f(a,x),b),b)	→	f(f(a,f(f(x,b),b)),b)	(2)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

f^#(a,f(a,f(a,f(x,b))))	→	f^#(f(a,f(a,f(a,x))),b)	(3)
f^#(a,f(a,f(a,f(x,b))))	→	f^#(a,f(a,f(a,x)))	(4)
f^#(a,f(a,f(a,f(x,b))))	→	f^#(a,f(a,x))	(5)
f^#(a,f(a,f(a,f(x,b))))	→	f^#(a,x)	(6)
f^#(f(f(a,x),b),b)	→	f^#(f(a,f(f(x,b),b)),b)	(7)
f^#(f(f(a,x),b),b)	→	f^#(a,f(f(x,b),b))	(8)
f^#(f(f(a,x),b),b)	→	f^#(f(x,b),b)	(9)
f^#(f(f(a,x),b),b)	→	f^#(x,b)	(10)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

f^#(a,f(a,f(a,f(x,b))))	→	f^#(a,f(a,x))	(5)
f^#(a,f(a,f(a,f(x,b))))	→	f^#(a,f(a,f(a,x)))	(4)
f^#(a,f(a,f(a,f(x,b))))	→	f^#(a,x)	(6)

1.1.1 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

[f(x₁, x₂)]	=	1 · x₁ + 1 · x₂
[a]	=	2
[b]	=	0
[f^#(x₁, x₂)]	=	1 · x₁ + 1 · x₂

the pairs

f^#(a,f(a,f(a,f(x,b))))	→	f^#(a,f(a,x))	(5)
f^#(a,f(a,f(a,f(x,b))))	→	f^#(a,x)	(6)

and no rules could be deleted.

1.1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals

[f^#(x₁, x₂)]

-∞

-∞	-∞	0
-∞	-∞	-∞
-∞	-∞	-∞

· x₁ +

-∞	-∞	0
-∞	-∞	-∞
-∞	-∞	-∞

· x₂

[a]

-∞

[f(x₁, x₂)]

-∞

1	-∞	-∞
-∞	-∞	-∞
-∞	-∞	0

· x₁ +

-∞	-∞	1
0	-∞	-∞
-∞	0	-∞

· x₂

[b]

the pair

f^#(a,f(a,f(a,f(x,b))))

→

f^#(a,f(a,f(a,x)))

(4)

could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.

The 2^nd component contains the pair

f^#(f(f(a,x),b),b)	→	f^#(f(x,b),b)	(9)
f^#(f(f(a,x),b),b)	→	f^#(f(a,f(f(x,b),b)),b)	(7)
f^#(f(f(a,x),b),b)	→	f^#(x,b)	(10)

1.1.2 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

[f(x₁, x₂)]	=	1 · x₁ + 2 · x₂
[a]	=	1
[b]	=	0
[f^#(x₁, x₂)]	=	2 · x₁ + 1 · x₂

the pairs

f^#(f(f(a,x),b),b)	→	f^#(f(x,b),b)	(9)
f^#(f(f(a,x),b),b)	→	f^#(x,b)	(10)

and no rules could be deleted.

1.1.2.1 Reduction Pair Processor

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

[f^#(x₁, x₂)]

0	1
0	0

· x₁ +

0	1
0	0

· x₂

[f(x₁, x₂)]

0	0
0	0

· x₁ +

0	1
1	0

· x₂

[a]

[b]

the pair

f^#(f(f(a,x),b),b)

→

f^#(f(a,f(f(x,b),b)),b)

(7)

could be deleted.

1.1.2.1.1 P is empty

There are no pairs anymore.