Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/161930)

The rewrite relation of the following TRS is considered.

There are 110 ruless (increase limit for explicit display).

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS

There are 110 ruless (increase limit for explicit display).

1.1 Closure Under Flat Contexts

Using the flat contexts

{0(☐), 1(☐), 2(☐)}

We obtain the transformed TRS

There are 200 ruless (increase limit for explicit display).

1.1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS

There are 600 ruless (increase limit for explicit display).

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals

[0₀(x₁)]	=	1 · x₁ + 1006
[0₁(x₁)]	=	1 · x₁ + 942
[1₁(x₁)]	=	1 · x₁ + 63
[1₀(x₁)]	=	1 · x₁ + 921
[0₂(x₁)]	=	1 · x₁
[1₂(x₁)]	=	1 · x₁
[2₀(x₁)]	=	1 · x₁ + 1954
[2₂(x₁)]	=	1 · x₁ + 932
[2₁(x₁)]	=	1 · x₁ + 1684

all of the following rules can be deleted.

There are 595 ruless (increase limit for explicit display).

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals

[1₀(x₁)]	=	1 · x₁
[0₀(x₁)]	=	1 · x₁
[0₂(x₁)]	=	1 · x₁ + 9
[2₁(x₁)]	=	1 · x₁ + 7
[0₁(x₁)]	=	1 · x₁
[1₁(x₁)]	=	1 · x₁
[1₂(x₁)]	=	1 · x₁
[2₂(x₁)]	=	1 · x₁ + 4
[2₀(x₁)]	=	1 · x₁

all of the following rules can be deleted.

1₀(0₀(0₂(2₁(1₀(0₁(1₁(x1)))))))	→	1₁(1₁(1₀(0₁(1₂(2₂(2₂(2₁(x1))))))))	(374)
1₂(2₀(0₂(2₂(2₂(2₂(2₂(2₂(2₁(1₂(2₁(1₀(0₀(0₀(0₂(2₂(2₀(0₂(2₂(2₂(x1))))))))))))))))))))	→	1₁(1₁(1₂(2₁(1₀(0₁(1₀(0₁(1₂(2₀(0₂(2₁(1₀(0₀(0₁(1₂(2₁(1₀(0₀(0₀(0₁(1₁(1₂(x1)))))))))))))))))))))))	(477)
2₂(2₀(0₀(0₂(2₂(2₀(0₂(2₁(1₂(2₁(1₂(2₂(2₀(0₂(2₀(0₁(1₀(0₀(0₁(1₁(1₀(0₀(0₀(0₀(0₂(x1)))))))))))))))))))))))))	→	2₁(1₁(1₁(1₁(1₂(2₀(0₂(2₂(2₀(0₂(2₀(0₀(0₀(0₁(1₂(2₁(1₂(2₀(0₀(0₂(2₁(1₂(2₀(0₂(2₀(0₁(1₂(x1)))))))))))))))))))))))))))	(501)
2₂(2₁(1₀(0₁(1₂(2₁(1₀(0₁(1₂(2₂(2₀(0₂(2₁(1₀(0₀(0₁(1₂(2₁(1₂(2₀(0₂(2₁(1₀(0₀(0₁(1₂(2₁(x1)))))))))))))))))))))))))))	→	2₀(0₁(1₁(1₂(2₀(0₂(2₂(2₂(2₂(2₀(0₂(2₁(1₀(0₁(1₂(2₂(2₁(1₁(1₂(2₂(2₂(2₂(2₁(1₁(1₂(2₀(0₀(0₀(0₁(x1)))))))))))))))))))))))))))))	(530)
0₂(2₀(0₀(0₀(0₀(0₀(0₂(2₀(0₂(2₁(1₁(1₁(x1))))))))))))	→	0₁(1₁(1₁(1₁(1₂(2₁(1₁(1₀(0₁(1₁(1₁(1₁(1₁(1₀(0₂(2₂(2₀(0₁(x1))))))))))))))))))	(659)

1.1.1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.