Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-78)

The relative rewrite relation R/S is considered where R is the following TRS

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

and S is the following TRS.

a(b(c(x1))) → c(c(b(x1))) (3)
b(a(c(x1))) → c(a(c(x1))) (4)
a(a(c(x1))) → a(a(b(x1))) (5)
b(c(a(x1))) → a(c(b(x1))) (6)
a(a(a(x1))) → c(a(b(x1))) (7)

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
[c(x1)] = x1 +
0
[b(x1)] = x1 +
0
[a(x1)] = x1 +
1
all of the following rules can be deleted.
a(b(c(x1))) → b(b(c(x1))) (2)
a(b(c(x1))) → c(c(b(x1))) (3)
a(a(a(x1))) → c(a(b(x1))) (7)

1.1 Split

We split R in the relative problem D/R-D and R-D, where the rules D

b(c(a(x1))) → a(c(b(x1))) (6)
are deleted.

1.1.1 Closure Under Flat Contexts

Using the flat contexts

{c(☐), b(☐), a(☐)}

We obtain the transformed TRS
c(b(c(a(x1)))) → c(a(c(b(x1)))) (8)
b(b(c(a(x1)))) → b(a(c(b(x1)))) (9)
a(b(c(a(x1)))) → a(a(c(b(x1)))) (10)
c(a(c(b(x1)))) → c(b(a(b(x1)))) (11)
c(b(a(c(x1)))) → c(c(a(c(x1)))) (12)
c(a(a(c(x1)))) → c(a(a(b(x1)))) (13)
b(a(c(b(x1)))) → b(b(a(b(x1)))) (14)
b(b(a(c(x1)))) → b(c(a(c(x1)))) (15)
b(a(a(c(x1)))) → b(a(a(b(x1)))) (16)
a(a(c(b(x1)))) → a(b(a(b(x1)))) (17)
a(b(a(c(x1)))) → a(c(a(c(x1)))) (18)
a(a(a(c(x1)))) → a(a(a(b(x1)))) (19)

1.1.1.1 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
a1(b0(c2(a2(x1)))) → a2(a0(c1(b2(x1)))) (20)
a1(b0(c2(a0(x1)))) → a2(a0(c1(b0(x1)))) (21)
a1(b0(c2(a1(x1)))) → a2(a0(c1(b1(x1)))) (22)
c1(b0(c2(a2(x1)))) → c2(a0(c1(b2(x1)))) (23)
c1(b0(c2(a0(x1)))) → c2(a0(c1(b0(x1)))) (24)
c1(b0(c2(a1(x1)))) → c2(a0(c1(b1(x1)))) (25)
b1(b0(c2(a2(x1)))) → b2(a0(c1(b2(x1)))) (26)
b1(b0(c2(a0(x1)))) → b2(a0(c1(b0(x1)))) (27)
b1(b0(c2(a1(x1)))) → b2(a0(c1(b1(x1)))) (28)
a2(a0(c1(b2(x1)))) → a1(b2(a1(b2(x1)))) (29)
a2(a0(c1(b0(x1)))) → a1(b2(a1(b0(x1)))) (30)
a2(a0(c1(b1(x1)))) → a1(b2(a1(b1(x1)))) (31)
c2(a0(c1(b2(x1)))) → c1(b2(a1(b2(x1)))) (32)
c2(a0(c1(b0(x1)))) → c1(b2(a1(b0(x1)))) (33)
c2(a0(c1(b1(x1)))) → c1(b2(a1(b1(x1)))) (34)
b2(a0(c1(b2(x1)))) → b1(b2(a1(b2(x1)))) (35)
b2(a0(c1(b0(x1)))) → b1(b2(a1(b0(x1)))) (36)
b2(a0(c1(b1(x1)))) → b1(b2(a1(b1(x1)))) (37)
a1(b2(a0(c2(x1)))) → a0(c2(a0(c2(x1)))) (38)
a1(b2(a0(c0(x1)))) → a0(c2(a0(c0(x1)))) (39)
a1(b2(a0(c1(x1)))) → a0(c2(a0(c1(x1)))) (40)
c1(b2(a0(c2(x1)))) → c0(c2(a0(c2(x1)))) (41)
c1(b2(a0(c0(x1)))) → c0(c2(a0(c0(x1)))) (42)
c1(b2(a0(c1(x1)))) → c0(c2(a0(c1(x1)))) (43)
b1(b2(a0(c2(x1)))) → b0(c2(a0(c2(x1)))) (44)
b1(b2(a0(c0(x1)))) → b0(c2(a0(c0(x1)))) (45)
b1(b2(a0(c1(x1)))) → b0(c2(a0(c1(x1)))) (46)
a2(a2(a0(c2(x1)))) → a2(a2(a1(b2(x1)))) (47)
a2(a2(a0(c0(x1)))) → a2(a2(a1(b0(x1)))) (48)
a2(a2(a0(c1(x1)))) → a2(a2(a1(b1(x1)))) (49)
c2(a2(a0(c2(x1)))) → c2(a2(a1(b2(x1)))) (50)
c2(a2(a0(c0(x1)))) → c2(a2(a1(b0(x1)))) (51)
c2(a2(a0(c1(x1)))) → c2(a2(a1(b1(x1)))) (52)
b2(a2(a0(c2(x1)))) → b2(a2(a1(b2(x1)))) (53)
b2(a2(a0(c0(x1)))) → b2(a2(a1(b0(x1)))) (54)
b2(a2(a0(c1(x1)))) → b2(a2(a1(b1(x1)))) (55)

1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the naturals
[c0(x1)] =
2
· x1 +
1
[c1(x1)] =
2
· x1 +
1
[c2(x1)] =
1
· x1 +
1
[b0(x1)] =
2
· x1 +
0
[b1(x1)] =
2
· x1 +
0
[b2(x1)] =
1
· x1 +
1
[a0(x1)] =
2
· x1 +
0
[a1(x1)] =
2
· x1 +
0
[a2(x1)] =
1
· x1 +
1
all of the following rules can be deleted.
a1(b0(c2(a2(x1)))) → a2(a0(c1(b2(x1)))) (20)
a1(b0(c2(a0(x1)))) → a2(a0(c1(b0(x1)))) (21)
a1(b0(c2(a1(x1)))) → a2(a0(c1(b1(x1)))) (22)
c1(b0(c2(a2(x1)))) → c2(a0(c1(b2(x1)))) (23)
c1(b0(c2(a0(x1)))) → c2(a0(c1(b0(x1)))) (24)
c1(b0(c2(a1(x1)))) → c2(a0(c1(b1(x1)))) (25)
b1(b0(c2(a2(x1)))) → b2(a0(c1(b2(x1)))) (26)
b1(b0(c2(a0(x1)))) → b2(a0(c1(b0(x1)))) (27)
b1(b0(c2(a1(x1)))) → b2(a0(c1(b1(x1)))) (28)
a2(a0(c1(b2(x1)))) → a1(b2(a1(b2(x1)))) (29)
a2(a0(c1(b0(x1)))) → a1(b2(a1(b0(x1)))) (30)
a2(a0(c1(b1(x1)))) → a1(b2(a1(b1(x1)))) (31)
b2(a0(c1(b2(x1)))) → b1(b2(a1(b2(x1)))) (35)
b2(a0(c1(b0(x1)))) → b1(b2(a1(b0(x1)))) (36)
b2(a0(c1(b1(x1)))) → b1(b2(a1(b1(x1)))) (37)
a2(a2(a0(c0(x1)))) → a2(a2(a1(b0(x1)))) (48)
a2(a2(a0(c1(x1)))) → a2(a2(a1(b1(x1)))) (49)
c2(a2(a0(c0(x1)))) → c2(a2(a1(b0(x1)))) (51)
c2(a2(a0(c1(x1)))) → c2(a2(a1(b1(x1)))) (52)
b2(a2(a0(c0(x1)))) → b2(a2(a1(b0(x1)))) (54)
b2(a2(a0(c1(x1)))) → b2(a2(a1(b1(x1)))) (55)

1.1.1.1.1.1 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 +
1
[c2(x1)] = x1 +
0
[b0(x1)] = x1 +
0
[b1(x1)] = x1 +
1
[b2(x1)] = x1 +
0
[a0(x1)] = x1 +
0
[a1(x1)] = x1 +
0
[a2(x1)] = x1 +
0
all of the following rules can be deleted.
c1(b2(a0(c2(x1)))) → c0(c2(a0(c2(x1)))) (41)
c1(b2(a0(c0(x1)))) → c0(c2(a0(c0(x1)))) (42)
c1(b2(a0(c1(x1)))) → c0(c2(a0(c1(x1)))) (43)
b1(b2(a0(c2(x1)))) → b0(c2(a0(c2(x1)))) (44)
b1(b2(a0(c0(x1)))) → b0(c2(a0(c0(x1)))) (45)
b1(b2(a0(c1(x1)))) → b0(c2(a0(c1(x1)))) (46)

1.1.1.1.1.1.1 R is empty

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

1.1.2 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(c(a(x1))) → b(a(b(x1))) (56)
c(a(b(x1))) → c(a(c(x1))) (57)
c(a(a(x1))) → b(a(a(x1))) (58)

1.1.2.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the naturals
[c(x1)] =
4
· x1 +
1
[b(x1)] =
4
· x1 +
1
[a(x1)] =
8
· x1 +
3
all of the following rules can be deleted.
b(c(a(x1))) → b(a(b(x1))) (56)

1.1.2.1.1 R is empty

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