YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

a(s(x0)) → s(a(x0))
b(a(b(s(x0)))) → a(b(s(a(x0))))
b(a(b(b(x0)))) → c(s(x0))
c(s(x0)) → a(b(a(b(x0))))
a(b(a(a(x0)))) → b(a(b(a(x0))))

Proof

1 String Reversal

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

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a(x1)] = 2 · x1 + 2
[c(x1)] = 8 · x1 + 15
[s(x1)] = 2 · x1 + 0
[b(x1)] = 2 · x1 + 2
the rules
b(b(a(b(x0)))) → s(c(x0))
s(c(x0)) → b(a(b(a(x0))))
a(a(b(a(x0)))) → a(b(a(b(x0))))
remain.

1.1.1 Bounds

The given TRS is match-bounded by 1. This is shown by the following automaton.