YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 Rule Removal

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

1.1 String Reversal

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

1.1.1 Bounds

The given TRS is match-bounded by 2. This is shown by the following automaton.