YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 Rule Removal

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

1.1 String Reversal

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[d(x1)] =
1 1
0 0
· x1 +
0 0
0 0
[a(x1)] =
1 0
3 1
· x1 +
0 0
1 0
[b(x1)] =
1 0
0 0
· x1 +
0 0
0 0
[c(x1)] =
1 0
0 0
· x1 +
0 0
0 0
the rules
b(d(b(x0))) → b(d(c(x0)))
b(b(b(x0))) → c(b(a(x0)))
c(d(x0)) → d(b(x0))
c(d(x0)) → d(b(d(x0)))
c(a(d(x0))) → b(b(x0))
remain.

1.1.1.1 Bounds

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