YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

0(0(1(2(x0)))) → 2(3(0(x0)))
4(2(5(2(x0)))) → 1(4(0(x0)))
4(5(3(4(x0)))) → 4(4(2(4(x0))))
1(3(5(5(2(x0))))) → 1(3(2(3(x0))))
5(2(2(1(2(x0))))) → 0(0(2(3(x0))))
2(2(5(3(2(2(x0)))))) → 5(1(1(0(3(x0)))))
2(5(1(2(1(1(x0)))))) → 5(2(1(2(4(1(x0))))))
3(4(1(4(2(4(x0)))))) → 1(3(3(3(x0))))
3(5(2(2(4(5(x0)))))) → 3(2(4(3(0(x0)))))
5(2(1(0(1(5(x0)))))) → 5(4(2(4(5(1(x0))))))
1(3(5(4(1(2(2(x0))))))) → 3(3(3(4(4(0(x0))))))
4(5(4(3(0(5(1(x0))))))) → 4(3(3(5(4(1(x0))))))
2(1(5(2(1(3(4(4(x0)))))))) → 4(0(3(4(0(1(2(x0)))))))
5(4(0(2(2(4(0(4(x0)))))))) → 3(1(5(1(3(0(4(x0)))))))
3(4(2(1(1(2(2(5(4(x0))))))))) → 3(3(3(1(3(3(4(x0)))))))
5(4(4(5(0(1(4(5(4(x0))))))))) → 1(5(5(0(4(1(4(5(4(x0)))))))))
5(2(1(3(1(5(2(5(4(4(x0)))))))))) → 5(3(4(5(0(1(4(0(3(x0)))))))))
2(4(1(2(5(2(4(1(3(2(0(3(x0)))))))))))) → 4(3(0(4(2(3(4(3(4(2(0(x0)))))))))))
0(2(3(5(4(2(2(1(0(3(3(5(0(x0))))))))))))) → 3(3(1(2(3(0(4(0(0(0(2(0(x0))))))))))))
2(1(0(2(1(4(0(0(2(0(0(0(5(2(x0)))))))))))))) → 2(2(4(2(1(4(3(0(5(1(3(3(0(x0)))))))))))))
4(5(0(3(1(3(2(2(5(2(2(4(1(3(2(x0))))))))))))))) → 4(5(3(2(1(4(5(0(0(0(4(5(4(0(0(x0)))))))))))))))
5(2(2(5(2(4(4(1(2(0(1(1(0(1(1(x0))))))))))))))) → 5(0(5(4(3(2(1(0(3(3(5(0(4(1(x0))))))))))))))
5(4(2(0(3(3(0(0(4(0(3(2(0(5(1(x0))))))))))))))) → 5(0(1(0(0(2(1(1(0(3(2(2(1(3(x0))))))))))))))
5(4(1(5(1(5(4(4(2(2(0(4(3(1(5(4(4(3(1(x0))))))))))))))))))) → 3(0(4(5(1(1(3(5(3(4(4(4(5(1(4(3(3(1(x0))))))))))))))))))
2(0(3(0(2(2(2(0(1(4(2(1(0(4(4(3(3(1(4(4(x0)))))))))))))))))))) → 5(2(4(1(1(4(5(1(0(1(2(0(3(0(1(2(3(4(3(1(x0))))))))))))))))))))

Proof

1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight function
prec(4) = 0 weight(4) = 1
prec(5) = 1 weight(5) = 1
prec(3) = 0 weight(3) = 1
prec(0) = 1 weight(0) = 1
prec(1) = 0 weight(1) = 1
prec(2) = 2 weight(2) = 1
all rules could be removed.

1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.