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

Proof

1 String Reversal

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

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a(x1)] = 8 · x1 + 1
[b(x1)] = 1 · x1 + 0
[c(x1)] = 1 · x1 + 3
the rules
b(a(x0)) → a(b(b(x0)))
c(b(x0)) → b(b(c(x0)))
remain.

1.1.1 String Reversal

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

1.1.1.1 Rule Removal

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

1.1.1.1.1 Bounds

The given TRS is match-bounded by 0. This is shown by the following automaton.