Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z045)

The rewrite relation of the following TRS is considered.

c(a(b(a(b(x1))))) → a(b(a(b(b(a(b(b(c(a(b(c(a(x1))))))))))))) (1)

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(☐), a(☐)}

We obtain the transformed TRS
c(c(a(b(a(b(x1)))))) → c(a(b(a(b(b(a(b(b(c(a(b(c(a(x1)))))))))))))) (2)
b(c(a(b(a(b(x1)))))) → b(a(b(a(b(b(a(b(b(c(a(b(c(a(x1)))))))))))))) (3)
a(c(a(b(a(b(x1)))))) → a(a(b(a(b(b(a(b(b(c(a(b(c(a(x1)))))))))))))) (4)

1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,1,2}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 3):

[c(x1)] = 3x1 + 0
[b(x1)] = 3x1 + 1
[a(x1)] = 3x1 + 2

We obtain the labeled TRS
c0(c2(a1(b2(a1(b0(x1)))))) → c2(a1(b2(a1(b1(b2(a1(b1(b0(c2(a1(b0(c2(a0(x1)))))))))))))) (5)
c0(c2(a1(b2(a1(b2(x1)))))) → c2(a1(b2(a1(b1(b2(a1(b1(b0(c2(a1(b0(c2(a2(x1)))))))))))))) (6)
c0(c2(a1(b2(a1(b1(x1)))))) → c2(a1(b2(a1(b1(b2(a1(b1(b0(c2(a1(b0(c2(a1(x1)))))))))))))) (7)
a0(c2(a1(b2(a1(b0(x1)))))) → a2(a1(b2(a1(b1(b2(a1(b1(b0(c2(a1(b0(c2(a0(x1)))))))))))))) (8)
a0(c2(a1(b2(a1(b2(x1)))))) → a2(a1(b2(a1(b1(b2(a1(b1(b0(c2(a1(b0(c2(a2(x1)))))))))))))) (9)
a0(c2(a1(b2(a1(b1(x1)))))) → a2(a1(b2(a1(b1(b2(a1(b1(b0(c2(a1(b0(c2(a1(x1)))))))))))))) (10)
b0(c2(a1(b2(a1(b0(x1)))))) → b2(a1(b2(a1(b1(b2(a1(b1(b0(c2(a1(b0(c2(a0(x1)))))))))))))) (11)
b0(c2(a1(b2(a1(b2(x1)))))) → b2(a1(b2(a1(b1(b2(a1(b1(b0(c2(a1(b0(c2(a2(x1)))))))))))))) (12)
b0(c2(a1(b2(a1(b1(x1)))))) → b2(a1(b2(a1(b1(b2(a1(b1(b0(c2(a1(b0(c2(a1(x1)))))))))))))) (13)

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
[c2(x1)] = x1 +
0
[b0(x1)] = x1 +
0
[b1(x1)] = x1 +
0
[b2(x1)] = x1 +
0
[a0(x1)] = x1 +
0
[a1(x1)] = x1 +
0
[a2(x1)] = x1 +
0
all of the following rules can be deleted.
c0(c2(a1(b2(a1(b0(x1)))))) → c2(a1(b2(a1(b1(b2(a1(b1(b0(c2(a1(b0(c2(a0(x1)))))))))))))) (5)
c0(c2(a1(b2(a1(b2(x1)))))) → c2(a1(b2(a1(b1(b2(a1(b1(b0(c2(a1(b0(c2(a2(x1)))))))))))))) (6)
c0(c2(a1(b2(a1(b1(x1)))))) → c2(a1(b2(a1(b1(b2(a1(b1(b0(c2(a1(b0(c2(a1(x1)))))))))))))) (7)

1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b0#(c2(a1(b2(a1(b0(x1)))))) → b0#(c2(a0(x1))) (14)
b0#(c2(a1(b2(a1(b0(x1)))))) → b0#(c2(a1(b0(c2(a0(x1)))))) (15)
b0#(c2(a1(b2(a1(b0(x1)))))) → a0#(x1) (16)
b0#(c2(a1(b2(a1(b1(x1)))))) → b0#(c2(a1(x1))) (17)
b0#(c2(a1(b2(a1(b1(x1)))))) → b0#(c2(a1(b0(c2(a1(x1)))))) (18)
b0#(c2(a1(b2(a1(b2(x1)))))) → b0#(c2(a1(b0(c2(a2(x1)))))) (19)
b0#(c2(a1(b2(a1(b2(x1)))))) → b0#(c2(a2(x1))) (20)
a0#(c2(a1(b2(a1(b0(x1)))))) → b0#(c2(a0(x1))) (21)
a0#(c2(a1(b2(a1(b0(x1)))))) → b0#(c2(a1(b0(c2(a0(x1)))))) (22)
a0#(c2(a1(b2(a1(b0(x1)))))) → a0#(x1) (23)
a0#(c2(a1(b2(a1(b1(x1)))))) → b0#(c2(a1(x1))) (24)
a0#(c2(a1(b2(a1(b1(x1)))))) → b0#(c2(a1(b0(c2(a1(x1)))))) (25)
a0#(c2(a1(b2(a1(b2(x1)))))) → b0#(c2(a1(b0(c2(a2(x1)))))) (26)
a0#(c2(a1(b2(a1(b2(x1)))))) → b0#(c2(a2(x1))) (27)

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.