YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[f(x1)] = 1 · x1 + 2
[a(x1)] = 1 · x1 + 4
[c(x1)] = 1 · x1 + 0
[b(x1)] = 4 · x1 + 0
[d(x1)] = 1 · x1 + 1
the rules
a(b(x0)) → b(d(x0))
b(d(x0)) → a(c(b(x0)))
c(f(x0)) → d(d(c(x0)))
d(d(x0)) → f(x0)
f(f(x0)) → a(x0)
remain.

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[f(x1)] = 4 · x1 + -∞
[a(x1)] = 2 · x1 + -∞
[c(x1)] = 0 · x1 + -∞
[b(x1)] = 0 · x1 + -∞
[d(x1)] = 2 · x1 + -∞
the rules
a(b(x0)) → b(d(x0))
b(d(x0)) → a(c(b(x0)))
c(f(x0)) → d(d(c(x0)))
d(d(x0)) → f(x0)
remain.

1.1.1 String Reversal

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

1.1.1.1 Bounds

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