YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

3(1(x0)) → 4(1(x0))
5(9(x0)) → 2(6(5(x0)))
3(5(x0)) → 8(9(7(x0)))
9(x0) → 3(2(3(x0)))
8(4(x0)) → 6(x0)
2(6(x0)) → 4(3(x0))
3(8(x0)) → 3(2(7(x0)))
9(x0) → 5(0(2(x0)))
8(8(4(x0))) → 1(9(x0))
7(1(x0)) → 6(9(x0))
3(9(x0)) → 9(3(x0))
7(5(x0)) → 1(0(x0))

Proof

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
1(3(x0)) → 1(4(x0))
9(5(x0)) → 5(6(2(x0)))
5(3(x0)) → 7(9(8(x0)))
9(x0) → 3(2(3(x0)))
4(8(x0)) → 6(x0)
6(2(x0)) → 3(4(x0))
8(3(x0)) → 7(2(3(x0)))
9(x0) → 2(0(5(x0)))
4(8(8(x0))) → 9(1(x0))
1(7(x0)) → 9(6(x0))
9(3(x0)) → 3(9(x0))
5(7(x0)) → 0(1(x0))

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[5(x1)] =
1 1 0
1 0 1
1 0 1
· x1 +
0 0 0
1 0 0
1 0 0
[8(x1)] =
1 1 0
0 0 0
1 0 1
· x1 +
0 0 0
0 0 0
1 0 0
[3(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[9(x1)] =
1 1 0
0 0 1
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[6(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[1(x1)] =
1 0 1
0 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[7(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[0(x1)] =
1 0 0
0 1 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[2(x1)] =
1 0 0
0 0 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[4(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
the rules
1(3(x0)) → 1(4(x0))
5(3(x0)) → 7(9(8(x0)))
9(x0) → 3(2(3(x0)))
6(2(x0)) → 3(4(x0))
8(3(x0)) → 7(2(3(x0)))
9(x0) → 2(0(5(x0)))
1(7(x0)) → 9(6(x0))
9(3(x0)) → 3(9(x0))
5(7(x0)) → 0(1(x0))
remain.

1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[5(x1)] = 1 · x1 + -∞
[8(x1)] = 0 · x1 + -∞
[3(x1)] = 0 · x1 + -∞
[9(x1)] = 1 · x1 + -∞
[6(x1)] = 0 · x1 + -∞
[1(x1)] = 1 · x1 + -∞
[7(x1)] = 0 · x1 + -∞
[0(x1)] = 0 · x1 + -∞
[2(x1)] = 0 · x1 + -∞
[4(x1)] = 0 · x1 + -∞
the rules
1(3(x0)) → 1(4(x0))
5(3(x0)) → 7(9(8(x0)))
6(2(x0)) → 3(4(x0))
8(3(x0)) → 7(2(3(x0)))
9(x0) → 2(0(5(x0)))
1(7(x0)) → 9(6(x0))
9(3(x0)) → 3(9(x0))
5(7(x0)) → 0(1(x0))
remain.

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[5(x1)] = 0 · x1 + -∞
[8(x1)] = 1 · x1 + -∞
[3(x1)] = 2 · x1 + -∞
[9(x1)] = 0 · x1 + -∞
[6(x1)] = 2 · x1 + -∞
[1(x1)] = 1 · x1 + -∞
[7(x1)] = 1 · x1 + -∞
[0(x1)] = 0 · x1 + -∞
[2(x1)] = 0 · x1 + -∞
[4(x1)] = 0 · x1 + -∞
the rules
5(3(x0)) → 7(9(8(x0)))
6(2(x0)) → 3(4(x0))
8(3(x0)) → 7(2(3(x0)))
9(x0) → 2(0(5(x0)))
1(7(x0)) → 9(6(x0))
9(3(x0)) → 3(9(x0))
5(7(x0)) → 0(1(x0))
remain.

1.1.1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
3(5(x0)) → 8(9(7(x0)))
2(6(x0)) → 4(3(x0))
3(8(x0)) → 3(2(7(x0)))
9(x0) → 5(0(2(x0)))
7(1(x0)) → 6(9(x0))
3(9(x0)) → 9(3(x0))
7(5(x0)) → 1(0(x0))

1.1.1.1.1.1 Bounds

The given TRS is match-bounded by 3. This is shown by the following automaton.