YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

a(a(b(x0))) → b(a(a(a(x0))))
b(a(b(a(x0)))) → a(b(b(x0)))

Proof

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(a(a(x0))) → a(a(a(b(x0))))
a(b(a(b(x0)))) → b(b(a(x0)))

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(a(a(x0))) → b#(x0)
b#(a(a(x0))) → a#(b(x0))
b#(a(a(x0))) → a#(a(b(x0)))
b#(a(a(x0))) → a#(a(a(b(x0))))
a#(b(a(b(x0)))) → a#(x0)
a#(b(a(b(x0)))) → b#(a(x0))
a#(b(a(b(x0)))) → b#(b(a(x0)))

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the arctic semiring over the integers
[a(x1)] = 0 · x1 + -∞
[a#(x1)] = 0 · x1 + 0
[b(x1)] = 8 · x1 + 0
[b#(x1)] = 8 · x1 + 0
together with the usable rules
b(a(a(x0))) → a(a(a(b(x0))))
a(b(a(b(x0)))) → b(b(a(x0)))
(w.r.t. the implicit argument filter of the reduction pair), the pairs
b#(a(a(x0))) → b#(x0)
b#(a(a(x0))) → a#(b(x0))
b#(a(a(x0))) → a#(a(b(x0)))
b#(a(a(x0))) → a#(a(a(b(x0))))
a#(b(a(b(x0)))) → b#(b(a(x0)))
remain.

1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[a(x1)] =
1 0
0 1
· x1 +
2 0
1 0
[a#(x1)] =
0 1
0 0
· x1 +
1 0
0 0
[b(x1)] =
1 1
1 0
· x1 +
0 0
0 0
[b#(x1)] =
1 0
0 0
· x1 +
0 0
0 0
the pair
a#(b(a(b(x0)))) → b#(b(a(x0)))
remains.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.