Certification Problem

Input (TPDB SRS_Standard/Waldmann_06_SRS/sym-4)

The rewrite relation of the following TRS is considered.

b(b(x1)) → c(c(c(c(x1)))) (1)
c(x1) → x1 (2)
b(c(b(x1))) → b(b(b(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

{c(☐), b(☐)}

We obtain the transformed TRS
c(b(b(x1))) → c(c(c(c(c(x1))))) (4)
c(c(x1)) → c(x1) (5)
c(b(c(b(x1)))) → c(b(b(b(x1)))) (6)
b(b(b(x1))) → b(c(c(c(c(x1))))) (7)
b(c(x1)) → b(x1) (8)
b(b(c(b(x1)))) → b(b(b(b(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):

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

We obtain the labeled TRS
b1(b1(b1(x1))) → b0(c0(c0(c0(c1(x1))))) (10)
b1(b1(b0(x1))) → b0(c0(c0(c0(c0(x1))))) (11)
c1(b1(b1(x1))) → c0(c0(c0(c0(c1(x1))))) (12)
c1(b1(b0(x1))) → c0(c0(c0(c0(c0(x1))))) (13)
b0(c1(x1)) → b1(x1) (14)
b0(c0(x1)) → b0(x1) (15)
c0(c1(x1)) → c1(x1) (16)
c0(c0(x1)) → c0(x1) (17)
b1(b0(c1(b1(x1)))) → b1(b1(b1(b1(x1)))) (18)
b1(b0(c1(b0(x1)))) → b1(b1(b1(b0(x1)))) (19)
c1(b0(c1(b1(x1)))) → c1(b1(b1(b1(x1)))) (20)
c1(b0(c1(b0(x1)))) → c1(b1(b1(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
[c0(x1)] = x1 +
1
[c1(x1)] = x1 +
7
[b0(x1)] = x1 +
7
[b1(x1)] = x1 +
6
all of the following rules can be deleted.
b1(b1(b1(x1))) → b0(c0(c0(c0(c1(x1))))) (10)
b1(b1(b0(x1))) → b0(c0(c0(c0(c0(x1))))) (11)
c1(b1(b1(x1))) → c0(c0(c0(c0(c1(x1))))) (12)
c1(b1(b0(x1))) → c0(c0(c0(c0(c0(x1))))) (13)
b0(c1(x1)) → b1(x1) (14)
b0(c0(x1)) → b0(x1) (15)
c0(c1(x1)) → c1(x1) (16)
c0(c0(x1)) → c0(x1) (17)
b1(b0(c1(b1(x1)))) → b1(b1(b1(b1(x1)))) (18)
b1(b0(c1(b0(x1)))) → b1(b1(b1(b0(x1)))) (19)
c1(b0(c1(b1(x1)))) → c1(b1(b1(b1(x1)))) (20)
c1(b0(c1(b0(x1)))) → c1(b1(b1(b0(x1)))) (21)

1.1.1.1 R is empty

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