Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z086)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(a(x1)) c#(b(x1)) (4)
a#(a(x1)) b#(x1) (5)
b#(b(x1)) c#(a(x1)) (6)
b#(b(x1)) a#(x1) (7)
c#(c(x1)) b#(a(x1)) (8)
c#(c(x1)) a#(x1) (9)

1.1 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a(x1)] = 1 + 2 · x1
[c(x1)] = 1 + 2 · x1
[b(x1)] = 1 + 2 · x1
[a#(x1)] = 3 · x1
[c#(x1)] = 2 · x1
[b#(x1)] = 2 · x1
the pairs
a#(a(x1)) c#(b(x1)) (4)
a#(a(x1)) b#(x1) (5)
b#(b(x1)) a#(x1) (7)
c#(c(x1)) a#(x1) (9)
and no rules could be deleted.

1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
[b#(x1)] =
-∞
-∞
-∞
+
0 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b(x1)] =
1
0
0
+
1 0 0
0 -∞ -∞
1 0 -∞
· x1
[c#(x1)] =
1
-∞
-∞
+
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
0
1
0
+
-∞ 0 -∞
1 0 0
0 0 0
· x1
[c(x1)] =
-∞
0
-∞
+
0 1 0
0 0 0
0 1 0
· x1
the pair
c#(c(x1)) b#(a(x1)) (8)
could be deleted.

1.1.1.1 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a(x1)] = 2 · x1
[c(x1)] = 2 · x1
[b(x1)] = 2 · x1
[b#(x1)] = 1 + 2 · x1
[c#(x1)] = 2 · x1
the pair
b#(b(x1)) c#(a(x1)) (6)
and no rules could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.