YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

f(f(x0)) → b(b(b(x0)))
a(f(x0)) → f(a(a(x0)))
b(b(x0)) → c(c(a(c(x0))))
d(b(x0)) → d(a(b(x0)))
c(c(x0)) → d(d(d(x0)))
b(d(x0)) → d(b(x0))
c(d(d(x0))) → f(x0)

Proof

1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[d(x1)] = 4 · x1 + -∞
[b(x1)] = 9 · x1 + -∞
[c(x1)] = 6 · x1 + -∞
[f(x1)] = 14 · x1 + -∞
[a(x1)] = 0 · x1 + -∞
the rules
a(f(x0)) → f(a(a(x0)))
b(b(x0)) → c(c(a(c(x0))))
d(b(x0)) → d(a(b(x0)))
c(c(x0)) → d(d(d(x0)))
b(d(x0)) → d(b(x0))
c(d(d(x0))) → f(x0)
remain.

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[d(x1)] = 0 · x1 + -∞
[b(x1)] = 8 · x1 + -∞
[c(x1)] = 3 · x1 + -∞
[f(x1)] = 0 · x1 + -∞
[a(x1)] = 0 · x1 + -∞
the rules
a(f(x0)) → f(a(a(x0)))
d(b(x0)) → d(a(b(x0)))
b(d(x0)) → d(b(x0))
remain.

1.1.1 String Reversal

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

1.1.1.1 Bounds

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