Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/25192)

The rewrite relation of the following TRS is considered.

0(x1) 1(x1) (1)
4(5(4(5(x1)))) 4(4(5(5(x1)))) (2)
5(5(5(5(5(5(4(4(4(4(4(4(x1)))))))))))) 2(x1) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[5(x1)] = x1 +
1
[4(x1)] = x1 +
1
[2(x1)] = x1 +
0
[1(x1)] = x1 +
0
[0(x1)] = x1 +
1
all of the following rules can be deleted.
0(x1) 1(x1) (1)
5(5(5(5(5(5(4(4(4(4(4(4(x1)))))))))))) 2(x1) (3)

1.1 Closure Under Flat Contexts

Using the flat contexts

{5(), 4()}

We obtain the transformed TRS
5(4(5(4(5(x1))))) 5(4(4(5(5(x1))))) (4)
4(4(5(4(5(x1))))) 4(4(4(5(5(x1))))) (5)

1.1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,1}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 2):

[5(x1)] = 2x1 + 0
[4(x1)] = 2x1 + 1

We obtain the labeled TRS
41(40(51(40(51(x1))))) 41(41(40(50(51(x1))))) (6)
41(40(51(40(50(x1))))) 41(41(40(50(50(x1))))) (7)
51(40(51(40(51(x1))))) 51(41(40(50(51(x1))))) (8)
51(40(51(40(50(x1))))) 51(41(40(50(50(x1))))) (9)

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[50(x1)] = x1 +
0
[51(x1)] = x1 +
1
[40(x1)] = x1 +
1
[41(x1)] = x1 +
0
all of the following rules can be deleted.
41(40(51(40(51(x1))))) 41(41(40(50(51(x1))))) (6)
41(40(51(40(50(x1))))) 41(41(40(50(50(x1))))) (7)
51(40(51(40(51(x1))))) 51(41(40(50(51(x1))))) (8)
51(40(51(40(50(x1))))) 51(41(40(50(50(x1))))) (9)

1.1.1.1.1 R is empty

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