Certification Problem

Input (TPDB SRS_Standard/Waldmann_06_SRS/sym-1)

The rewrite relation of the following TRS is considered.

a(a(x1)) → a(b(a(x1))) (1)
b(b(x1)) → a(a(x1)) (2)
a(b(b(a(x1)))) → x1 (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{b(☐), a(☐)}

We obtain the transformed TRS
b(a(a(x1))) → b(a(b(a(x1)))) (4)
b(b(b(x1))) → b(a(a(x1))) (5)
b(a(b(b(a(x1))))) → b(x1) (6)
a(a(a(x1))) → a(a(b(a(x1)))) (7)
a(b(b(x1))) → a(a(a(x1))) (8)
a(a(b(b(a(x1))))) → a(x1) (9)

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):

[b(x1)] = 2x1 + 0
[a(x1)] = 2x1 + 1

We obtain the labeled TRS
a1(a1(a1(x1))) → a1(a0(b1(a1(x1)))) (10)
a1(a1(a0(x1))) → a1(a0(b1(a0(x1)))) (11)
b1(a1(a1(x1))) → b1(a0(b1(a1(x1)))) (12)
b1(a1(a0(x1))) → b1(a0(b1(a0(x1)))) (13)
a0(b0(b1(x1))) → a1(a1(a1(x1))) (14)
a0(b0(b0(x1))) → a1(a1(a0(x1))) (15)
b0(b0(b1(x1))) → b1(a1(a1(x1))) (16)
b0(b0(b0(x1))) → b1(a1(a0(x1))) (17)
a1(a0(b0(b1(a1(x1))))) → a1(x1) (18)
a1(a0(b0(b1(a0(x1))))) → a0(x1) (19)
b1(a0(b0(b1(a1(x1))))) → b1(x1) (20)
b1(a0(b0(b1(a0(x1))))) → b0(x1) (21)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[b0(x1)] = x1 +
8
[b1(x1)] = x1 +
1
[a0(x1)] = x1 +
1
[a1(x1)] = x1 +
3
all of the following rules can be deleted.
a1(a1(a1(x1))) → a1(a0(b1(a1(x1)))) (10)
a1(a1(a0(x1))) → a1(a0(b1(a0(x1)))) (11)
b1(a1(a1(x1))) → b1(a0(b1(a1(x1)))) (12)
b1(a1(a0(x1))) → b1(a0(b1(a0(x1)))) (13)
a0(b0(b1(x1))) → a1(a1(a1(x1))) (14)
a0(b0(b0(x1))) → a1(a1(a0(x1))) (15)
b0(b0(b1(x1))) → b1(a1(a1(x1))) (16)
b0(b0(b0(x1))) → b1(a1(a0(x1))) (17)
a1(a0(b0(b1(a1(x1))))) → a1(x1) (18)
a1(a0(b0(b1(a0(x1))))) → a0(x1) (19)
b1(a0(b0(b1(a1(x1))))) → b1(x1) (20)
b1(a0(b0(b1(a0(x1))))) → b0(x1) (21)

1.1.1.1 R is empty

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