Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-92)

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

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

and S is the following TRS.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{c(☐), a(☐), b(☐)}

We obtain the transformed TRS
c(a(a(x1))) → c(c(b(x1))) (1)
c(a(b(x1))) → c(b(a(x1))) (2)
b(b(a(x1))) → b(b(c(x1))) (3)
c(b(c(b(x1)))) → c(a(a(a(x1)))) (7)
a(b(c(b(x1)))) → a(a(a(a(x1)))) (8)
b(b(c(b(x1)))) → b(a(a(a(x1)))) (9)
a(b(c(x1))) → a(a(c(x1))) (5)
c(c(a(a(x1)))) → c(a(b(a(x1)))) (10)
a(c(a(a(x1)))) → a(a(b(a(x1)))) (11)
b(c(a(a(x1)))) → b(a(b(a(x1)))) (12)

1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
ca(aa(ac(x1))) → cc(cb(bc(x1))) (13)
ca(aa(aa(x1))) → cc(cb(ba(x1))) (14)
ca(aa(ab(x1))) → cc(cb(bb(x1))) (15)
ca(ab(bc(x1))) → cb(ba(ac(x1))) (16)
ca(ab(ba(x1))) → cb(ba(aa(x1))) (17)
ca(ab(bb(x1))) → cb(ba(ab(x1))) (18)
bb(ba(ac(x1))) → bb(bc(cc(x1))) (19)
bb(ba(aa(x1))) → bb(bc(ca(x1))) (20)
bb(ba(ab(x1))) → bb(bc(cb(x1))) (21)
cb(bc(cb(bc(x1)))) → ca(aa(aa(ac(x1)))) (22)
cb(bc(cb(ba(x1)))) → ca(aa(aa(aa(x1)))) (23)
cb(bc(cb(bb(x1)))) → ca(aa(aa(ab(x1)))) (24)
ab(bc(cb(bc(x1)))) → aa(aa(aa(ac(x1)))) (25)
ab(bc(cb(ba(x1)))) → aa(aa(aa(aa(x1)))) (26)
ab(bc(cb(bb(x1)))) → aa(aa(aa(ab(x1)))) (27)
bb(bc(cb(bc(x1)))) → ba(aa(aa(ac(x1)))) (28)
bb(bc(cb(ba(x1)))) → ba(aa(aa(aa(x1)))) (29)
bb(bc(cb(bb(x1)))) → ba(aa(aa(ab(x1)))) (30)
ab(bc(cc(x1))) → aa(ac(cc(x1))) (31)
ab(bc(ca(x1))) → aa(ac(ca(x1))) (32)
ab(bc(cb(x1))) → aa(ac(cb(x1))) (33)
cc(ca(aa(ac(x1)))) → ca(ab(ba(ac(x1)))) (34)
cc(ca(aa(aa(x1)))) → ca(ab(ba(aa(x1)))) (35)
cc(ca(aa(ab(x1)))) → ca(ab(ba(ab(x1)))) (36)
ac(ca(aa(ac(x1)))) → aa(ab(ba(ac(x1)))) (37)
ac(ca(aa(aa(x1)))) → aa(ab(ba(aa(x1)))) (38)
ac(ca(aa(ab(x1)))) → aa(ab(ba(ab(x1)))) (39)
bc(ca(aa(ac(x1)))) → ba(ab(ba(ac(x1)))) (40)
bc(ca(aa(aa(x1)))) → ba(ab(ba(aa(x1)))) (41)
bc(ca(aa(ab(x1)))) → ba(ab(ba(ab(x1)))) (42)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[ca(x1)] =
0
0
+
1 0
0 0
· x1
[aa(x1)] =
0
0
+
1 0
0 0
· x1
[ac(x1)] =
0
0
+
1 0
0 0
· x1
[cc(x1)] =
0
0
+
1 0
0 0
· x1
[cb(x1)] =
0
0
+
1 0
0 0
· x1
[bc(x1)] =
0
0
+
1 0
0 0
· x1
[ba(x1)] =
0
0
+
1 0
0 0
· x1
[ab(x1)] =
0
0
+
1 2
0 0
· x1
[bb(x1)] =
0
2
+
1 2
0 0
· x1
all of the following rules can be deleted.
ca(ab(bb(x1))) → cb(ba(ab(x1))) (18)

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[ca(x1)] =
0
0
+
2 0
0 0
· x1
[aa(x1)] =
0
0
+
2 0
0 0
· x1
[ac(x1)] =
0
0
+
2 2
0 0
· x1
[cc(x1)] =
0
0
+
2 0
0 0
· x1
[cb(x1)] =
0
0
+
2 0
0 1
· x1
[bc(x1)] =
0
0
+
2 2
0 0
· x1
[ba(x1)] =
0
0
+
2 0
0 0
· x1
[ab(x1)] =
0
0
+
2 1
0 0
· x1
[bb(x1)] =
0
2
+
2 0
0 2
· x1
all of the following rules can be deleted.
cb(bc(cb(bb(x1)))) → ca(aa(aa(ab(x1)))) (24)
ab(bc(cb(bb(x1)))) → aa(aa(aa(ab(x1)))) (27)
bb(bc(cb(bb(x1)))) → ba(aa(aa(ab(x1)))) (30)

1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[ca(x1)] =
0
0
+
1 0
0 2
· x1
[aa(x1)] =
0
1
+
1 0
0 0
· x1
[ac(x1)] =
0
1
+
1 2
1 0
· x1
[cc(x1)] =
0
2
+
1 0
0 0
· x1
[cb(x1)] =
0
2
+
1 2
0 0
· x1
[bc(x1)] =
0
0
+
1 0
0 1
· x1
[ba(x1)] =
0
0
+
1 0
0 0
· x1
[ab(x1)] =
0
1
+
1 2
0 0
· x1
[bb(x1)] =
0
0
+
1 0
0 0
· x1
all of the following rules can be deleted.
cb(bc(cb(bc(x1)))) → ca(aa(aa(ac(x1)))) (22)
cb(bc(cb(ba(x1)))) → ca(aa(aa(aa(x1)))) (23)
ab(bc(cb(bc(x1)))) → aa(aa(aa(ac(x1)))) (25)
ab(bc(cb(ba(x1)))) → aa(aa(aa(aa(x1)))) (26)
ac(ca(aa(ac(x1)))) → aa(ab(ba(ac(x1)))) (37)
ac(ca(aa(aa(x1)))) → aa(ab(ba(aa(x1)))) (38)
ac(ca(aa(ab(x1)))) → aa(ab(ba(ab(x1)))) (39)

1.1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[ca(x1)] =
0
2
+
2 0
1 0
· x1
[aa(x1)] =
0
0
+
2 0
0 0
· x1
[ac(x1)] =
0
0
+
1 2
0 0
· x1
[cc(x1)] =
0
0
+
1 2
0 2
· x1
[cb(x1)] =
0
0
+
2 2
0 0
· x1
[bc(x1)] =
0
0
+
2 0
0 2
· x1
[ba(x1)] =
0
0
+
2 0
0 0
· x1
[ab(x1)] =
0
0
+
2 2
0 0
· x1
[bb(x1)] =
0
0
+
2 0
0 0
· x1
all of the following rules can be deleted.
cc(ca(aa(ac(x1)))) → ca(ab(ba(ac(x1)))) (34)
cc(ca(aa(aa(x1)))) → ca(ab(ba(aa(x1)))) (35)
cc(ca(aa(ab(x1)))) → ca(ab(ba(ab(x1)))) (36)

1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[ca(x1)] = 1 + 1 · x1
[aa(x1)] = 1 + 1 · x1
[ac(x1)] = 1 · x1
[cc(x1)] = 1 · x1
[cb(x1)] = 1 + 1 · x1
[bc(x1)] = 1 + 1 · x1
[ba(x1)] = 1 + 1 · x1
[ab(x1)] = 1 + 1 · x1
[bb(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
ca(aa(aa(x1))) → cc(cb(ba(x1))) (14)
ca(aa(ab(x1))) → cc(cb(bb(x1))) (15)
ca(ab(bc(x1))) → cb(ba(ac(x1))) (16)
bb(bc(cb(bc(x1)))) → ba(aa(aa(ac(x1)))) (28)
ab(bc(cc(x1))) → aa(ac(cc(x1))) (31)
ab(bc(ca(x1))) → aa(ac(ca(x1))) (32)
ab(bc(cb(x1))) → aa(ac(cb(x1))) (33)

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(ca) = 3 weight(ca) = 6
prec(aa) = 5 weight(aa) = 7
prec(ac) = 1 weight(ac) = 2
prec(cc) = 4 weight(cc) = 1
prec(cb) = 2 weight(cb) = 3
prec(bc) = 7 weight(bc) = 7
prec(ab) = 8 weight(ab) = 4
prec(ba) = 6 weight(ba) = 8
prec(bb) = 0 weight(bb) = 12
all of the following rules can be deleted.
ca(aa(ac(x1))) → cc(cb(bc(x1))) (13)
ca(ab(ba(x1))) → cb(ba(aa(x1))) (17)
bb(ba(ac(x1))) → bb(bc(cc(x1))) (19)
bb(ba(aa(x1))) → bb(bc(ca(x1))) (20)
bb(ba(ab(x1))) → bb(bc(cb(x1))) (21)
bb(bc(cb(ba(x1)))) → ba(aa(aa(aa(x1)))) (29)
bc(ca(aa(ac(x1)))) → ba(ab(ba(ac(x1)))) (40)
bc(ca(aa(aa(x1)))) → ba(ab(ba(aa(x1)))) (41)
bc(ca(aa(ab(x1)))) → ba(ab(ba(ab(x1)))) (42)

1.1.1.1.1.1.1.1.1 R is empty

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