YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 String Reversal

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

1.1 Rule Removal

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

1.1.1 String Reversal

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

1.1.1.1 Bounds

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