Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z039)

The rewrite relation of the following TRS is considered.

c(a(a(b(c(a(x1)))))) → a(a(b(c(c(a(a(b(c(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(a(b(c(a(x1))))))) → c(a(a(b(c(c(a(a(b(c(x1)))))))))) (2)
b(c(a(a(b(c(a(x1))))))) → b(a(a(b(c(c(a(a(b(c(x1)))))))))) (3)
a(c(a(a(b(c(a(x1))))))) → a(a(a(b(c(c(a(a(b(c(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(a2(a1(b0(c2(a0(x1))))))) → c2(a2(a1(b0(c0(c2(a2(a1(b0(c0(x1)))))))))) (5)
c0(c2(a2(a1(b0(c2(a2(x1))))))) → c2(a2(a1(b0(c0(c2(a2(a1(b0(c2(x1)))))))))) (6)
c0(c2(a2(a1(b0(c2(a1(x1))))))) → c2(a2(a1(b0(c0(c2(a2(a1(b0(c1(x1)))))))))) (7)
a0(c2(a2(a1(b0(c2(a0(x1))))))) → a2(a2(a1(b0(c0(c2(a2(a1(b0(c0(x1)))))))))) (8)
a0(c2(a2(a1(b0(c2(a2(x1))))))) → a2(a2(a1(b0(c0(c2(a2(a1(b0(c2(x1)))))))))) (9)
a0(c2(a2(a1(b0(c2(a1(x1))))))) → a2(a2(a1(b0(c0(c2(a2(a1(b0(c1(x1)))))))))) (10)
b0(c2(a2(a1(b0(c2(a0(x1))))))) → b2(a2(a1(b0(c0(c2(a2(a1(b0(c0(x1)))))))))) (11)
b0(c2(a2(a1(b0(c2(a2(x1))))))) → b2(a2(a1(b0(c0(c2(a2(a1(b0(c2(x1)))))))))) (12)
b0(c2(a2(a1(b0(c2(a1(x1))))))) → b2(a2(a1(b0(c0(c2(a2(a1(b0(c1(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 +
0
[c1(x1)] = x1 +
0
[c2(x1)] = x1 +
0
[b0(x1)] = x1 +
0
[b2(x1)] = x1 +
0
[a0(x1)] = x1 +
1
[a1(x1)] = x1 +
0
[a2(x1)] = x1 +
0
all of the following rules can be deleted.
c0(c2(a2(a1(b0(c2(a0(x1))))))) → c2(a2(a1(b0(c0(c2(a2(a1(b0(c0(x1)))))))))) (5)
a0(c2(a2(a1(b0(c2(a0(x1))))))) → a2(a2(a1(b0(c0(c2(a2(a1(b0(c0(x1)))))))))) (8)
a0(c2(a2(a1(b0(c2(a2(x1))))))) → a2(a2(a1(b0(c0(c2(a2(a1(b0(c2(x1)))))))))) (9)
a0(c2(a2(a1(b0(c2(a1(x1))))))) → a2(a2(a1(b0(c0(c2(a2(a1(b0(c1(x1)))))))))) (10)
b0(c2(a2(a1(b0(c2(a0(x1))))))) → b2(a2(a1(b0(c0(c2(a2(a1(b0(c0(x1)))))))))) (11)

1.1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a2(c2(b0(a1(a2(c2(c0(x1))))))) → c2(b0(a1(a2(c2(c0(b0(a1(a2(c2(x1)))))))))) (14)
a1(c2(b0(a1(a2(c2(c0(x1))))))) → c1(b0(a1(a2(c2(c0(b0(a1(a2(c2(x1)))))))))) (15)
a2(c2(b0(a1(a2(c2(b0(x1))))))) → c2(b0(a1(a2(c2(c0(b0(a1(a2(b2(x1)))))))))) (16)
a1(c2(b0(a1(a2(c2(b0(x1))))))) → c1(b0(a1(a2(c2(c0(b0(a1(a2(b2(x1)))))))))) (17)

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a1#(c2(b0(a1(a2(c2(c0(x1))))))) → a1#(a2(c2(x1))) (18)
a1#(c2(b0(a1(a2(c2(c0(x1))))))) → a1#(a2(c2(c0(b0(a1(a2(c2(x1)))))))) (19)
a1#(c2(b0(a1(a2(c2(c0(x1))))))) → a2#(c2(x1)) (20)
a1#(c2(b0(a1(a2(c2(c0(x1))))))) → a2#(c2(c0(b0(a1(a2(c2(x1))))))) (21)
a1#(c2(b0(a1(a2(c2(b0(x1))))))) → a1#(a2(c2(c0(b0(a1(a2(b2(x1)))))))) (22)
a1#(c2(b0(a1(a2(c2(b0(x1))))))) → a1#(a2(b2(x1))) (23)
a1#(c2(b0(a1(a2(c2(b0(x1))))))) → a2#(c2(c0(b0(a1(a2(b2(x1))))))) (24)
a1#(c2(b0(a1(a2(c2(b0(x1))))))) → a2#(b2(x1)) (25)
a2#(c2(b0(a1(a2(c2(c0(x1))))))) → a1#(a2(c2(x1))) (26)
a2#(c2(b0(a1(a2(c2(c0(x1))))))) → a1#(a2(c2(c0(b0(a1(a2(c2(x1)))))))) (27)
a2#(c2(b0(a1(a2(c2(c0(x1))))))) → a2#(c2(x1)) (28)
a2#(c2(b0(a1(a2(c2(c0(x1))))))) → a2#(c2(c0(b0(a1(a2(c2(x1))))))) (29)
a2#(c2(b0(a1(a2(c2(b0(x1))))))) → a1#(a2(c2(c0(b0(a1(a2(b2(x1)))))))) (30)
a2#(c2(b0(a1(a2(c2(b0(x1))))))) → a1#(a2(b2(x1))) (31)
a2#(c2(b0(a1(a2(c2(b0(x1))))))) → a2#(c2(c0(b0(a1(a2(b2(x1))))))) (32)
a2#(c2(b0(a1(a2(c2(b0(x1))))))) → a2#(b2(x1)) (33)

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.