Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-39)

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

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

and S is the following TRS.

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

{a(☐), c(☐), b(☐)}

We obtain the transformed TRS
a(a(c(x1))) → a(c(c(x1))) (1)
a(b(b(c(x1)))) → a(a(b(b(x1)))) (7)
c(b(b(c(x1)))) → c(a(b(b(x1)))) (8)
b(b(b(c(x1)))) → b(a(b(b(x1)))) (9)
b(b(a(x1))) → b(a(c(x1))) (5)
a(b(a(c(x1)))) → a(c(a(c(x1)))) (10)
c(b(a(c(x1)))) → c(c(a(c(x1)))) (11)
b(b(a(c(x1)))) → b(c(a(c(x1)))) (12)
a(c(b(c(x1)))) → a(b(b(b(x1)))) (13)
c(c(b(c(x1)))) → c(b(b(b(x1)))) (14)
b(c(b(c(x1)))) → b(b(b(b(x1)))) (15)
a(c(c(b(x1)))) → a(a(c(a(x1)))) (16)
c(c(c(b(x1)))) → c(a(c(a(x1)))) (17)
b(c(c(b(x1)))) → b(a(c(a(x1)))) (18)

1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
aa(ac(ca(x1))) → ac(cc(ca(x1))) (19)
aa(ac(cc(x1))) → ac(cc(cc(x1))) (20)
aa(ac(cb(x1))) → ac(cc(cb(x1))) (21)
ab(bb(bc(ca(x1)))) → aa(ab(bb(ba(x1)))) (22)
ab(bb(bc(cc(x1)))) → aa(ab(bb(bc(x1)))) (23)
ab(bb(bc(cb(x1)))) → aa(ab(bb(bb(x1)))) (24)
cb(bb(bc(ca(x1)))) → ca(ab(bb(ba(x1)))) (25)
cb(bb(bc(cc(x1)))) → ca(ab(bb(bc(x1)))) (26)
cb(bb(bc(cb(x1)))) → ca(ab(bb(bb(x1)))) (27)
bb(bb(bc(ca(x1)))) → ba(ab(bb(ba(x1)))) (28)
bb(bb(bc(cc(x1)))) → ba(ab(bb(bc(x1)))) (29)
bb(bb(bc(cb(x1)))) → ba(ab(bb(bb(x1)))) (30)
bb(ba(aa(x1))) → ba(ac(ca(x1))) (31)
bb(ba(ac(x1))) → ba(ac(cc(x1))) (32)
bb(ba(ab(x1))) → ba(ac(cb(x1))) (33)
ab(ba(ac(ca(x1)))) → ac(ca(ac(ca(x1)))) (34)
ab(ba(ac(cc(x1)))) → ac(ca(ac(cc(x1)))) (35)
ab(ba(ac(cb(x1)))) → ac(ca(ac(cb(x1)))) (36)
cb(ba(ac(ca(x1)))) → cc(ca(ac(ca(x1)))) (37)
cb(ba(ac(cc(x1)))) → cc(ca(ac(cc(x1)))) (38)
cb(ba(ac(cb(x1)))) → cc(ca(ac(cb(x1)))) (39)
bb(ba(ac(ca(x1)))) → bc(ca(ac(ca(x1)))) (40)
bb(ba(ac(cc(x1)))) → bc(ca(ac(cc(x1)))) (41)
bb(ba(ac(cb(x1)))) → bc(ca(ac(cb(x1)))) (42)
ac(cb(bc(ca(x1)))) → ab(bb(bb(ba(x1)))) (43)
ac(cb(bc(cc(x1)))) → ab(bb(bb(bc(x1)))) (44)
ac(cb(bc(cb(x1)))) → ab(bb(bb(bb(x1)))) (45)
cc(cb(bc(ca(x1)))) → cb(bb(bb(ba(x1)))) (46)
cc(cb(bc(cc(x1)))) → cb(bb(bb(bc(x1)))) (47)
cc(cb(bc(cb(x1)))) → cb(bb(bb(bb(x1)))) (48)
bc(cb(bc(ca(x1)))) → bb(bb(bb(ba(x1)))) (49)
bc(cb(bc(cc(x1)))) → bb(bb(bb(bc(x1)))) (50)
bc(cb(bc(cb(x1)))) → bb(bb(bb(bb(x1)))) (51)
ac(cc(cb(ba(x1)))) → aa(ac(ca(aa(x1)))) (52)
ac(cc(cb(bc(x1)))) → aa(ac(ca(ac(x1)))) (53)
ac(cc(cb(bb(x1)))) → aa(ac(ca(ab(x1)))) (54)
cc(cc(cb(ba(x1)))) → ca(ac(ca(aa(x1)))) (55)
cc(cc(cb(bc(x1)))) → ca(ac(ca(ac(x1)))) (56)
cc(cc(cb(bb(x1)))) → ca(ac(ca(ab(x1)))) (57)
bc(cc(cb(ba(x1)))) → ba(ac(ca(aa(x1)))) (58)
bc(cc(cb(bc(x1)))) → ba(ac(ca(ac(x1)))) (59)
bc(cc(cb(bb(x1)))) → ba(ac(ca(ab(x1)))) (60)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[aa(x1)] = 1 + 1 · x1
[ac(x1)] = 1 · x1
[ca(x1)] = 1 · x1
[cc(x1)] = 1 + 1 · x1
[cb(x1)] = 1 + 1 · x1
[ab(x1)] = 1 · x1
[bb(x1)] = 1 + 1 · x1
[bc(x1)] = 1 + 1 · x1
[ba(x1)] = 1 · x1
all of the following rules can be deleted.
cb(bb(bc(ca(x1)))) → ca(ab(bb(ba(x1)))) (25)
cb(bb(bc(cc(x1)))) → ca(ab(bb(bc(x1)))) (26)
cb(bb(bc(cb(x1)))) → ca(ab(bb(bb(x1)))) (27)
bb(bb(bc(ca(x1)))) → ba(ab(bb(ba(x1)))) (28)
bb(bb(bc(cc(x1)))) → ba(ab(bb(bc(x1)))) (29)
bb(bb(bc(cb(x1)))) → ba(ab(bb(bb(x1)))) (30)
bb(ba(aa(x1))) → ba(ac(ca(x1))) (31)
ac(cc(cb(bc(x1)))) → aa(ac(ca(ac(x1)))) (53)
ac(cc(cb(bb(x1)))) → aa(ac(ca(ab(x1)))) (54)
cc(cc(cb(ba(x1)))) → ca(ac(ca(aa(x1)))) (55)
cc(cc(cb(bc(x1)))) → ca(ac(ca(ac(x1)))) (56)
cc(cc(cb(bb(x1)))) → ca(ac(ca(ab(x1)))) (57)
bc(cc(cb(ba(x1)))) → ba(ac(ca(aa(x1)))) (58)
bc(cc(cb(bc(x1)))) → ba(ac(ca(ac(x1)))) (59)
bc(cc(cb(bb(x1)))) → ba(ac(ca(ab(x1)))) (60)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[aa(x1)] = 1 · x1
[ac(x1)] = 1 · x1
[ca(x1)] = 1 · x1
[cc(x1)] = 1 · x1
[cb(x1)] = 1 + 1 · x1
[ab(x1)] = 1 + 1 · x1
[bb(x1)] = 1 · x1
[bc(x1)] = 1 · x1
[ba(x1)] = 1 · x1
all of the following rules can be deleted.
ab(bb(bc(cb(x1)))) → aa(ab(bb(bb(x1)))) (24)
ab(ba(ac(ca(x1)))) → ac(ca(ac(ca(x1)))) (34)
ab(ba(ac(cc(x1)))) → ac(ca(ac(cc(x1)))) (35)
ab(ba(ac(cb(x1)))) → ac(ca(ac(cb(x1)))) (36)
cb(ba(ac(ca(x1)))) → cc(ca(ac(ca(x1)))) (37)
cb(ba(ac(cc(x1)))) → cc(ca(ac(cc(x1)))) (38)
cb(ba(ac(cb(x1)))) → cc(ca(ac(cb(x1)))) (39)
ac(cb(bc(cb(x1)))) → ab(bb(bb(bb(x1)))) (45)
cc(cb(bc(cb(x1)))) → cb(bb(bb(bb(x1)))) (48)
bc(cb(bc(ca(x1)))) → bb(bb(bb(ba(x1)))) (49)
bc(cb(bc(cc(x1)))) → bb(bb(bb(bc(x1)))) (50)
bc(cb(bc(cb(x1)))) → bb(bb(bb(bb(x1)))) (51)
ac(cc(cb(ba(x1)))) → aa(ac(ca(aa(x1)))) (52)

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
+
1 0
0 2
· x1
[ac(x1)] =
0
2
+
1 2
0 0
· x1
[ca(x1)] =
0
2
+
1 0
0 0
· x1
[cc(x1)] =
0
0
+
1 0
0 0
· x1
[cb(x1)] =
0
0
+
1 0
0 0
· x1
[ab(x1)] =
0
0
+
1 0
0 0
· x1
[bb(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
all of the following rules can be deleted.
aa(ac(ca(x1))) → ac(cc(ca(x1))) (19)

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
+
1 2
0 2
· x1
[ac(x1)] =
0
2
+
2 0
1 0
· x1
[cc(x1)] =
0
2
+
2 0
0 0
· x1
[cb(x1)] =
0
0
+
2 0
0 0
· x1
[ab(x1)] =
0
0
+
2 1
0 0
· x1
[bb(x1)] =
0
0
+
2 0
0 0
· x1
[bc(x1)] =
0
0
+
2 0
0 0
· x1
[ca(x1)] =
0
0
+
2 0
0 0
· x1
[ba(x1)] =
0
0
+
2 0
0 0
· x1
all of the following rules can be deleted.
aa(ac(cc(x1))) → ac(cc(cc(x1))) (20)
aa(ac(cb(x1))) → ac(cc(cb(x1))) (21)

1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[ab(x1)] = 1 · x1
[bb(x1)] = 1 + 1 · x1
[bc(x1)] = 1 · x1
[ca(x1)] = 1 + 1 · x1
[aa(x1)] = 1 · x1
[ba(x1)] = 1 · x1
[cc(x1)] = 1 + 1 · x1
[ac(x1)] = 1 + 1 · x1
[cb(x1)] = 1 · x1
all of the following rules can be deleted.
ab(bb(bc(ca(x1)))) → aa(ab(bb(ba(x1)))) (22)
ab(bb(bc(cc(x1)))) → aa(ab(bb(bc(x1)))) (23)

1.1.1.1.1.1.1.1 R is empty

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