NO Non-Confluence Proof

Non-Confluence Proof

by Hakusan

Input

The rewrite relation of the following TRS is considered.

0(0(1(0(x)))) → 0(2(0(0(3(1(x))))))
0(0(1(0(x)))) → 0(2(0(4(1(0(x))))))
0(0(1(0(x)))) → 2(0(0(0(2(1(x))))))
3(0(1(0(x)))) → 0(2(3(1(0(x)))))
3(0(1(0(x)))) → 3(1(0(0(2(x)))))
3(0(1(0(x)))) → 3(1(1(0(0(x)))))
3(0(1(0(x)))) → 3(1(2(0(0(x)))))
3(0(1(0(x)))) → 3(1(5(0(0(x)))))
3(0(1(0(x)))) → 3(5(1(0(0(x)))))
3(0(1(0(x)))) → 5(0(3(1(0(x)))))
3(0(1(0(x)))) → 2(0(2(3(1(0(x))))))
3(0(1(0(x)))) → 2(2(0(3(1(0(x))))))
3(0(1(0(x)))) → 3(1(5(0(0(0(x))))))
3(0(1(0(x)))) → 3(1(5(0(2(0(x))))))
3(0(1(0(x)))) → 3(1(5(1(0(0(x))))))
3(0(1(0(x)))) → 3(1(5(2(0(0(x))))))
3(0(1(0(x)))) → 3(1(5(5(0(0(x))))))
3(0(1(0(x)))) → 3(2(2(1(0(0(x))))))
3(0(1(0(x)))) → 3(5(1(0(0(2(x))))))
3(0(1(0(x)))) → 3(5(1(5(0(0(x))))))
3(0(1(0(x)))) → 5(1(1(3(0(0(x))))))
3(4(1(0(x)))) → 3(1(2(4(0(x)))))
3(4(1(0(x)))) → 3(1(4(0(2(x)))))
3(4(1(0(x)))) → 3(1(5(4(0(x)))))
3(4(1(0(x)))) → 3(4(2(1(0(x)))))
3(4(1(0(x)))) → 3(1(1(5(4(0(x))))))
3(4(1(0(x)))) → 3(1(2(1(4(0(x))))))
3(4(1(0(x)))) → 3(1(2(5(4(0(x))))))
3(4(1(0(x)))) → 3(1(4(2(0(2(x))))))
3(4(1(0(x)))) → 3(1(5(4(0(2(x))))))
3(4(1(0(x)))) → 3(1(5(5(4(0(x))))))
3(4(1(0(x)))) → 3(4(2(1(1(0(x))))))
3(4(1(0(x)))) → 3(4(5(1(2(0(x))))))
0(1(4(1(0(x))))) → 0(1(1(4(0(2(x))))))
0(2(0(1(0(x))))) → 0(2(0(0(3(1(x))))))
0(2(0(1(0(x))))) → 2(0(0(0(3(1(x))))))
0(3(0(1(0(x))))) → 0(0(3(1(3(0(x))))))
0(3(0(1(0(x))))) → 0(0(3(3(1(0(x))))))
0(3(0(1(0(x))))) → 0(0(3(5(1(0(x))))))
0(3(0(1(0(x))))) → 2(0(0(3(1(0(x))))))
0(3(4(1(0(x))))) → 0(2(0(4(3(1(x))))))
0(5(0(1(0(x))))) → 0(0(0(1(5(2(x))))))
0(5(0(1(0(x))))) → 0(0(1(5(1(0(x))))))
0(5(0(1(0(x))))) → 0(2(0(0(1(5(x))))))
3(0(1(0(0(x))))) → 3(1(3(0(0(0(x))))))
3(0(1(1(0(x))))) → 3(1(0(1(2(0(x))))))
3(0(2(1(0(x))))) → 2(0(3(1(1(0(x))))))
3(0(2(1(0(x))))) → 2(3(1(5(0(0(x))))))
3(0(2(1(0(x))))) → 3(1(2(0(1(0(x))))))
3(0(2(1(0(x))))) → 3(1(2(0(5(0(x))))))
3(0(5(1(0(x))))) → 3(1(5(2(0(0(x))))))
3(1(0(1(0(x))))) → 2(0(3(1(1(0(x))))))
3(1(0(1(0(x))))) → 3(1(1(1(0(0(x))))))
3(1(0(1(0(x))))) → 3(1(2(1(0(0(x))))))
3(1(4(1(0(x))))) → 3(1(2(1(4(0(x))))))
3(1(4(1(0(x))))) → 3(1(5(1(4(0(x))))))
3(2(0(1(0(x))))) → 0(2(3(1(5(0(x))))))
3(2(0(1(0(x))))) → 2(0(3(1(1(0(x))))))
3(3(0(1(0(x))))) → 3(1(2(0(3(0(x))))))
3(3(0(1(0(x))))) → 3(1(2(3(0(0(x))))))
3(3(4(1(0(x))))) → 3(1(2(4(3(0(x))))))
3(3(4(1(0(x))))) → 3(1(3(4(0(2(x))))))
3(3(4(1(0(x))))) → 3(1(4(3(1(0(x))))))
3(4(0(1(0(x))))) → 0(2(4(1(3(0(x))))))
3(4(0(1(0(x))))) → 3(1(4(0(0(2(x))))))
3(4(0(1(0(x))))) → 3(2(0(4(1(0(x))))))
3(4(4(1(0(x))))) → 3(1(1(4(4(0(x))))))

Proof

1 Non-Joinable Fork

The system is not confluent due to the following forking derivations.

t0 = 0(0(1(0(y1))))
→ε 0(2(0(4(1(0(y1))))))
= t1

t0 = 0(0(1(0(y1))))
→ε 0(2(0(0(3(1(y1))))))
= t1

The two resulting terms cannot be joined for the following reason:

Tool configuration

Hakusan