Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-79)

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

b(b(c(x1))) → b(c(c(x1))) (1)
a(a(a(x1))) → c(b(a(x1))) (2)
b(b(c(x1))) → b(a(a(x1))) (3)
c(b(c(x1))) → a(a(a(x1))) (4)

and S is the following TRS.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
c(b(b(x1))) → c(c(b(x1))) (8)
a(a(a(x1))) → a(b(c(x1))) (9)
c(b(b(x1))) → a(a(b(x1))) (10)
c(b(c(x1))) → a(a(a(x1))) (4)
a(a(b(x1))) → c(a(b(x1))) (11)
b(c(a(x1))) → c(a(b(x1))) (12)
b(b(a(x1))) → a(c(a(x1))) (13)

1.1 Closure Under Flat Contexts

Using the flat contexts

{c(☐), b(☐), a(☐)}

We obtain the transformed TRS
c(b(b(x1))) → c(c(b(x1))) (8)
a(a(a(x1))) → a(b(c(x1))) (9)
c(c(b(b(x1)))) → c(a(a(b(x1)))) (14)
b(c(b(b(x1)))) → b(a(a(b(x1)))) (15)
a(c(b(b(x1)))) → a(a(a(b(x1)))) (16)
c(c(b(c(x1)))) → c(a(a(a(x1)))) (17)
b(c(b(c(x1)))) → b(a(a(a(x1)))) (18)
a(c(b(c(x1)))) → a(a(a(a(x1)))) (19)
c(a(a(b(x1)))) → c(c(a(b(x1)))) (20)
b(a(a(b(x1)))) → b(c(a(b(x1)))) (21)
a(a(a(b(x1)))) → a(c(a(b(x1)))) (22)
c(b(c(a(x1)))) → c(c(a(b(x1)))) (23)
b(b(c(a(x1)))) → b(c(a(b(x1)))) (24)
a(b(c(a(x1)))) → a(c(a(b(x1)))) (25)
c(b(b(a(x1)))) → c(a(c(a(x1)))) (26)
b(b(b(a(x1)))) → b(a(c(a(x1)))) (27)
a(b(b(a(x1)))) → a(a(c(a(x1)))) (28)

1.1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
cb(bb(bc(x1))) → cc(cb(bc(x1))) (29)
cb(bb(bb(x1))) → cc(cb(bb(x1))) (30)
cb(bb(ba(x1))) → cc(cb(ba(x1))) (31)
aa(aa(ac(x1))) → ab(bc(cc(x1))) (32)
aa(aa(ab(x1))) → ab(bc(cb(x1))) (33)
aa(aa(aa(x1))) → ab(bc(ca(x1))) (34)
cc(cb(bb(bc(x1)))) → ca(aa(ab(bc(x1)))) (35)
cc(cb(bb(bb(x1)))) → ca(aa(ab(bb(x1)))) (36)
cc(cb(bb(ba(x1)))) → ca(aa(ab(ba(x1)))) (37)
bc(cb(bb(bc(x1)))) → ba(aa(ab(bc(x1)))) (38)
bc(cb(bb(bb(x1)))) → ba(aa(ab(bb(x1)))) (39)
bc(cb(bb(ba(x1)))) → ba(aa(ab(ba(x1)))) (40)
ac(cb(bb(bc(x1)))) → aa(aa(ab(bc(x1)))) (41)
ac(cb(bb(bb(x1)))) → aa(aa(ab(bb(x1)))) (42)
ac(cb(bb(ba(x1)))) → aa(aa(ab(ba(x1)))) (43)
cc(cb(bc(cc(x1)))) → ca(aa(aa(ac(x1)))) (44)
cc(cb(bc(cb(x1)))) → ca(aa(aa(ab(x1)))) (45)
cc(cb(bc(ca(x1)))) → ca(aa(aa(aa(x1)))) (46)
bc(cb(bc(cc(x1)))) → ba(aa(aa(ac(x1)))) (47)
bc(cb(bc(cb(x1)))) → ba(aa(aa(ab(x1)))) (48)
bc(cb(bc(ca(x1)))) → ba(aa(aa(aa(x1)))) (49)
ac(cb(bc(cc(x1)))) → aa(aa(aa(ac(x1)))) (50)
ac(cb(bc(cb(x1)))) → aa(aa(aa(ab(x1)))) (51)
ac(cb(bc(ca(x1)))) → aa(aa(aa(aa(x1)))) (52)
ca(aa(ab(bc(x1)))) → cc(ca(ab(bc(x1)))) (53)
ca(aa(ab(bb(x1)))) → cc(ca(ab(bb(x1)))) (54)
ca(aa(ab(ba(x1)))) → cc(ca(ab(ba(x1)))) (55)
ba(aa(ab(bc(x1)))) → bc(ca(ab(bc(x1)))) (56)
ba(aa(ab(bb(x1)))) → bc(ca(ab(bb(x1)))) (57)
ba(aa(ab(ba(x1)))) → bc(ca(ab(ba(x1)))) (58)
aa(aa(ab(bc(x1)))) → ac(ca(ab(bc(x1)))) (59)
aa(aa(ab(bb(x1)))) → ac(ca(ab(bb(x1)))) (60)
aa(aa(ab(ba(x1)))) → ac(ca(ab(ba(x1)))) (61)
cb(bc(ca(ac(x1)))) → cc(ca(ab(bc(x1)))) (62)
cb(bc(ca(ab(x1)))) → cc(ca(ab(bb(x1)))) (63)
cb(bc(ca(aa(x1)))) → cc(ca(ab(ba(x1)))) (64)
bb(bc(ca(ac(x1)))) → bc(ca(ab(bc(x1)))) (65)
bb(bc(ca(ab(x1)))) → bc(ca(ab(bb(x1)))) (66)
bb(bc(ca(aa(x1)))) → bc(ca(ab(ba(x1)))) (67)
ab(bc(ca(ac(x1)))) → ac(ca(ab(bc(x1)))) (68)
ab(bc(ca(ab(x1)))) → ac(ca(ab(bb(x1)))) (69)
ab(bc(ca(aa(x1)))) → ac(ca(ab(ba(x1)))) (70)
cb(bb(ba(ac(x1)))) → ca(ac(ca(ac(x1)))) (71)
cb(bb(ba(ab(x1)))) → ca(ac(ca(ab(x1)))) (72)
cb(bb(ba(aa(x1)))) → ca(ac(ca(aa(x1)))) (73)
bb(bb(ba(ac(x1)))) → ba(ac(ca(ac(x1)))) (74)
bb(bb(ba(ab(x1)))) → ba(ac(ca(ab(x1)))) (75)
bb(bb(ba(aa(x1)))) → ba(ac(ca(aa(x1)))) (76)
ab(bb(ba(ac(x1)))) → aa(ac(ca(ac(x1)))) (77)
ab(bb(ba(ab(x1)))) → aa(ac(ca(ab(x1)))) (78)
ab(bb(ba(aa(x1)))) → aa(ac(ca(aa(x1)))) (79)

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[cb(x1)] =
0
1
+
1 0
0 0
· x1
[bb(x1)] =
2
0
+
1 0
0 0
· x1
[bc(x1)] =
2
0
+
1 0
0 0
· x1
[cc(x1)] =
0
0
+
1 2
0 0
· x1
[ba(x1)] =
1
0
+
1 0
0 0
· x1
[aa(x1)] =
1
0
+
1 0
0 0
· x1
[ac(x1)] =
0
0
+
1 2
0 0
· x1
[ab(x1)] =
0
0
+
1 0
0 0
· x1
[ca(x1)] =
0
0
+
1 0
0 0
· x1
all of the following rules can be deleted.
aa(aa(aa(x1))) → ab(bc(ca(x1))) (34)
cc(cb(bb(bc(x1)))) → ca(aa(ab(bc(x1)))) (35)
cc(cb(bb(bb(x1)))) → ca(aa(ab(bb(x1)))) (36)
cc(cb(bb(ba(x1)))) → ca(aa(ab(ba(x1)))) (37)
bc(cb(bb(bc(x1)))) → ba(aa(ab(bc(x1)))) (38)
bc(cb(bb(bb(x1)))) → ba(aa(ab(bb(x1)))) (39)
bc(cb(bb(ba(x1)))) → ba(aa(ab(ba(x1)))) (40)
ac(cb(bb(bc(x1)))) → aa(aa(ab(bc(x1)))) (41)
ac(cb(bb(bb(x1)))) → aa(aa(ab(bb(x1)))) (42)
ac(cb(bb(ba(x1)))) → aa(aa(ab(ba(x1)))) (43)
cc(cb(bc(cc(x1)))) → ca(aa(aa(ac(x1)))) (44)
cc(cb(bc(cb(x1)))) → ca(aa(aa(ab(x1)))) (45)
cc(cb(bc(ca(x1)))) → ca(aa(aa(aa(x1)))) (46)
bc(cb(bc(cc(x1)))) → ba(aa(aa(ac(x1)))) (47)
bc(cb(bc(cb(x1)))) → ba(aa(aa(ab(x1)))) (48)
ac(cb(bc(cc(x1)))) → aa(aa(aa(ac(x1)))) (50)
ac(cb(bc(cb(x1)))) → aa(aa(aa(ab(x1)))) (51)
ca(aa(ab(bc(x1)))) → cc(ca(ab(bc(x1)))) (53)
ca(aa(ab(bb(x1)))) → cc(ca(ab(bb(x1)))) (54)
ca(aa(ab(ba(x1)))) → cc(ca(ab(ba(x1)))) (55)
aa(aa(ab(bc(x1)))) → ac(ca(ab(bc(x1)))) (59)
aa(aa(ab(bb(x1)))) → ac(ca(ab(bb(x1)))) (60)
aa(aa(ab(ba(x1)))) → ac(ca(ab(ba(x1)))) (61)
cb(bc(ca(aa(x1)))) → cc(ca(ab(ba(x1)))) (64)
bb(bc(ca(aa(x1)))) → bc(ca(ab(ba(x1)))) (67)
ab(bc(ca(aa(x1)))) → ac(ca(ab(ba(x1)))) (70)
cb(bb(ba(ac(x1)))) → ca(ac(ca(ac(x1)))) (71)
cb(bb(ba(ab(x1)))) → ca(ac(ca(ab(x1)))) (72)
cb(bb(ba(aa(x1)))) → ca(ac(ca(aa(x1)))) (73)
bb(bb(ba(ac(x1)))) → ba(ac(ca(ac(x1)))) (74)
bb(bb(ba(ab(x1)))) → ba(ac(ca(ab(x1)))) (75)
bb(bb(ba(aa(x1)))) → ba(ac(ca(aa(x1)))) (76)
ab(bb(ba(ac(x1)))) → aa(ac(ca(ac(x1)))) (77)
ab(bb(ba(ab(x1)))) → aa(ac(ca(ab(x1)))) (78)
ab(bb(ba(aa(x1)))) → aa(ac(ca(aa(x1)))) (79)

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[cb(x1)] = 1 · x1
[bb(x1)] = 1 · x1
[bc(x1)] = 1 · x1
[cc(x1)] = 1 · x1
[ba(x1)] = 1 + 1 · x1
[aa(x1)] = 1 · x1
[ac(x1)] = 1 · x1
[ab(x1)] = 1 · x1
[ca(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
ac(cb(bc(ca(x1)))) → aa(aa(aa(aa(x1)))) (52)

1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[cb(x1)] = 1 + 1 · x1
[bb(x1)] = 1 · x1
[bc(x1)] = 1 + 1 · x1
[cc(x1)] = 1 · x1
[ba(x1)] = 1 + 1 · x1
[aa(x1)] = 1 + 1 · x1
[ac(x1)] = 1 + 1 · x1
[ab(x1)] = 1 · x1
[ca(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
aa(aa(ac(x1))) → ab(bc(cc(x1))) (32)
cb(bc(ca(ac(x1)))) → cc(ca(ab(bc(x1)))) (62)
cb(bc(ca(ab(x1)))) → cc(ca(ab(bb(x1)))) (63)

1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[cb(x1)] = 1 · x1
[bb(x1)] = 1 + 1 · x1
[bc(x1)] = 1 · x1
[cc(x1)] = 1 · x1
[ba(x1)] = 1 · x1
[aa(x1)] = 1 · x1
[ab(x1)] = 1 + 1 · x1
[ca(x1)] = 1 · x1
[ac(x1)] = 1 · x1
all of the following rules can be deleted.
cb(bb(bc(x1))) → cc(cb(bc(x1))) (29)
cb(bb(bb(x1))) → cc(cb(bb(x1))) (30)
cb(bb(ba(x1))) → cc(cb(ba(x1))) (31)

1.1.1.1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[aa(x1)] =
0
0
+
2 0
2 0
· x1
[ab(x1)] =
2
0
+
2 0
0 0
· x1
[bc(x1)] =
0
0
+
2 2
0 0
· x1
[cb(x1)] =
0
0
+
1 0
1 0
· x1
[ca(x1)] =
0
0
+
2 0
2 0
· x1
[ba(x1)] =
0
0
+
2 2
0 0
· x1
[bb(x1)] =
1
0
+
2 0
0 0
· x1
[ac(x1)] =
2
0
+
2 2
0 0
· x1
all of the following rules can be deleted.
aa(aa(ab(x1))) → ab(bc(cb(x1))) (33)
bb(bc(ca(ac(x1)))) → bc(ca(ab(bc(x1)))) (65)
bb(bc(ca(ab(x1)))) → bc(ca(ab(bb(x1)))) (66)
ab(bc(ca(ac(x1)))) → ac(ca(ab(bc(x1)))) (68)

1.1.1.1.1.1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(bc) = 3 weight(bc) = 3
prec(cb) = 0 weight(cb) = 6
prec(ca) = 4 weight(ca) = 1
prec(ba) = 5 weight(ba) = 1
prec(aa) = 1 weight(aa) = 3
prec(ab) = 6 weight(ab) = 3
prec(bb) = 7 weight(bb) = 4
prec(ac) = 2 weight(ac) = 1
all of the following rules can be deleted.
bc(cb(bc(ca(x1)))) → ba(aa(aa(aa(x1)))) (49)
ba(aa(ab(bc(x1)))) → bc(ca(ab(bc(x1)))) (56)
ba(aa(ab(bb(x1)))) → bc(ca(ab(bb(x1)))) (57)
ba(aa(ab(ba(x1)))) → bc(ca(ab(ba(x1)))) (58)
ab(bc(ca(ab(x1)))) → ac(ca(ab(bb(x1)))) (69)

1.1.1.1.1.1.1.1.1.1 R is empty

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