Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-536)

The rewrite relation of the following TRS is considered.

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

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

1.1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1.1 Dependency Pair Transformation

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

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
[b#(x1)] =
0
-∞
-∞
+
-∞ -∞ 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
1
-∞
-∞
+
0 0 1
0 0 0
-∞ 0 0
· x1
[c(x1)] =
0
0
-∞
+
1 -∞ -∞
-∞ -∞ 0
0 0 0
· x1
[c#(x1)] =
0
-∞
-∞
+
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b(x1)] =
0
0
0
+
-∞ 0 -∞
-∞ 0 -∞
-∞ 0 -∞
· x1
the pair
c#(a(x1)) → b#(x1) (7)
could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.