Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size11/size-11-alpha-2-num-3)

The rewrite relation of the following TRS is considered.

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

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#(x1) → b#(x1) (3)
b#(a(a(b(x1)))) → a#(a(b(b(a(x1))))) (4)
b#(a(a(b(x1)))) → a#(b(b(a(x1)))) (5)
b#(a(a(b(x1)))) → b#(b(a(x1))) (6)
b#(a(a(b(x1)))) → b#(a(x1)) (7)
b#(a(a(b(x1)))) → a#(x1) (8)

1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
[a#(x1)] =
0
-∞
-∞
+
0 1 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b#(x1)] =
0
-∞
-∞
+
-∞ 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
0
0
0
+
-∞ 0 1
0 0 -∞
0 0 -∞
· x1
[b(x1)] =
0
0
0
+
-∞ 0 -∞
-∞ 0 -∞
0 0 -∞
· x1
the pairs
b#(a(a(b(x1)))) → b#(b(a(x1))) (6)
b#(a(a(b(x1)))) → b#(a(x1)) (7)
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
[a#(x1)] =
-∞
-∞
-∞
+
0 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b#(x1)] =
-∞
-∞
-∞
+
0 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
-∞
-∞
-∞
+
0 0 -∞
0 0 0
1 1 -∞
· x1
[b(x1)] =
-∞
-∞
-∞
+
0 -∞ -∞
0 -∞ -∞
1 1 -∞
· x1
the pairs
b#(a(a(b(x1)))) → a#(b(b(a(x1)))) (5)
b#(a(a(b(x1)))) → a#(x1) (8)
could be deleted.

1.1.1.1 Reduction Pair Processor

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

1.1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals
[b#(x1)] = 1 + 1 · x1
[a(x1)] = 1
[b(x1)] = 0
[a#(x1)] = 0
having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
b#(a(a(b(x1)))) → a#(a(b(b(a(x1))))) (4)
could be deleted.

1.1.1.1.1.1 P is empty

There are no pairs anymore.