YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 String Reversal

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

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[b(x1)] = 0 · x1 + -∞
[a(x1)] = 1 · x1 + -∞
the rules
a(b(a(x0))) → a(b(b(a(b(x0)))))
a(b(a(x0))) → b(a(b(b(a(x0)))))
remain.

1.1.1.1 String Reversal

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

1.1.1.1.1 Bounds

The given TRS is match-bounded by 0. This is shown by the following automaton.