YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

0(0(0(1(0(2(0(2(1(2(0(2(2(x0))))))))))))) 0(0(1(0(1(1(0(2(1(0(0(0(1(0(1(1(0(x0)))))))))))))))))
0(0(0(1(1(2(1(2(1(1(0(0(0(x0))))))))))))) 1(0(0(2(2(2(2(1(1(2(0(2(0(0(2(1(0(x0)))))))))))))))))
0(1(0(1(0(0(1(0(0(2(1(2(0(x0))))))))))))) 0(1(0(2(0(0(2(1(0(0(0(0(0(1(0(0(0(x0)))))))))))))))))
0(1(2(0(2(0(1(1(1(1(0(0(2(x0))))))))))))) 0(0(0(0(0(2(0(2(2(0(2(2(2(0(0(0(0(x0)))))))))))))))))
0(1(2(1(1(0(0(2(2(1(0(2(2(x0))))))))))))) 1(0(0(2(1(0(0(2(0(0(0(2(0(2(2(2(2(x0)))))))))))))))))
0(1(2(2(0(0(2(0(0(0(2(0(2(x0))))))))))))) 2(1(0(0(0(2(1(1(0(2(0(1(0(2(1(0(2(x0)))))))))))))))))
0(2(0(1(0(1(1(0(1(2(0(0(1(x0))))))))))))) 0(1(1(0(0(0(2(1(1(1(0(2(0(0(2(0(1(x0)))))))))))))))))
1(0(0(1(0(2(2(0(0(1(2(0(0(x0))))))))))))) 0(0(0(0(0(1(0(1(0(1(1(0(1(0(0(2(0(x0)))))))))))))))))
1(0(1(1(1(2(2(2(2(1(0(0(0(x0))))))))))))) 2(1(0(0(1(0(1(0(2(2(1(1(0(0(2(2(2(x0)))))))))))))))))
1(1(0(0(1(0(0(0(0(1(1(1(2(x0))))))))))))) 1(1(0(2(1(0(0(2(1(0(1(0(0(2(0(1(2(x0)))))))))))))))))
1(1(2(0(1(0(2(1(2(0(1(0(2(x0))))))))))))) 1(1(2(1(0(1(0(2(1(1(1(0(1(0(2(0(2(x0)))))))))))))))))
1(1(2(2(1(1(2(1(0(0(1(0(2(x0))))))))))))) 0(0(2(0(2(0(0(0(2(0(0(2(0(0(2(2(2(x0)))))))))))))))))
2(0(0(0(1(1(2(1(0(2(2(0(0(x0))))))))))))) 1(1(0(2(0(1(0(2(2(1(1(1(0(2(2(0(0(x0)))))))))))))))))
2(0(0(1(1(2(2(1(0(2(2(2(2(x0))))))))))))) 1(0(2(2(1(0(1(2(1(0(1(0(0(2(0(2(0(x0)))))))))))))))))
2(0(0(2(1(2(1(1(0(1(0(0(2(x0))))))))))))) 2(1(1(1(1(0(2(0(1(0(1(0(2(1(0(0(2(x0)))))))))))))))))
2(1(1(2(2(0(2(1(0(0(0(1(0(x0))))))))))))) 1(1(2(0(0(2(0(0(1(0(0(2(0(0(0(1(0(x0)))))))))))))))))
2(1(2(1(1(2(1(0(0(1(0(1(0(x0))))))))))))) 2(2(1(1(1(0(0(0(0(1(0(0(2(2(0(1(0(x0)))))))))))))))))
2(1(2(2(0(2(1(0(2(0(2(1(0(x0))))))))))))) 1(0(1(2(0(0(2(0(0(2(1(0(2(1(0(0(0(x0)))))))))))))))))
2(2(0(1(1(1(1(0(1(0(1(2(0(x0))))))))))))) 0(1(1(0(1(1(0(0(2(2(0(1(1(0(1(0(0(x0)))))))))))))))))

Proof

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
2(2(0(2(1(2(0(2(0(1(0(0(0(x0))))))))))))) 0(1(1(0(1(0(0(0(1(2(0(1(1(0(1(0(0(x0)))))))))))))))))
0(0(0(1(1(2(1(2(1(1(0(0(0(x0))))))))))))) 0(1(2(0(0(2(0(2(1(1(2(2(2(2(0(0(1(x0)))))))))))))))))
0(2(1(2(0(0(1(0(0(1(0(1(0(x0))))))))))))) 0(0(0(1(0(0(0(0(0(1(2(0(0(2(0(1(0(x0)))))))))))))))))
2(0(0(1(1(1(1(0(2(0(2(1(0(x0))))))))))))) 0(0(0(0(2(2(2(0(2(2(0(2(0(0(0(0(0(x0)))))))))))))))))
2(2(0(1(2(2(0(0(1(1(2(1(0(x0))))))))))))) 2(2(2(2(0(2(0(0(0(2(0(0(1(2(0(0(1(x0)))))))))))))))))
2(0(2(0(0(0(2(0(0(2(2(1(0(x0))))))))))))) 2(0(1(2(0(1(0(2(0(1(1(2(0(0(0(1(2(x0)))))))))))))))))
1(0(0(2(1(0(1(1(0(1(0(2(0(x0))))))))))))) 1(0(2(0(0(2(0(1(1(1(2(0(0(0(1(1(0(x0)))))))))))))))))
0(0(2(1(0(0(2(2(0(1(0(0(1(x0))))))))))))) 0(2(0(0(1(0(1(1(0(1(0(1(0(0(0(0(0(x0)))))))))))))))))
0(0(0(1(2(2(2(2(1(1(1(0(1(x0))))))))))))) 2(2(2(0(0(1(1(2(2(0(1(0(1(0(0(1(2(x0)))))))))))))))))
2(1(1(1(0(0(0(0(1(0(0(1(1(x0))))))))))))) 2(1(0(2(0(0(1(0(1(2(0(0(1(2(0(1(1(x0)))))))))))))))))
2(0(1(0(2(1(2(0(1(0(2(1(1(x0))))))))))))) 2(0(2(0(1(0(1(1(1(2(0(1(0(1(2(1(1(x0)))))))))))))))))
2(0(1(0(0(1(2(1(1(2(2(1(1(x0))))))))))))) 2(2(2(0(0(2(0(0(2(0(0(0(2(0(2(0(0(x0)))))))))))))))))
0(0(2(2(0(1(2(1(1(0(0(0(2(x0))))))))))))) 0(0(2(2(0(1(1(1(2(2(0(1(0(2(0(1(1(x0)))))))))))))))))
2(2(2(2(0(1(2(2(1(1(0(0(2(x0))))))))))))) 0(2(0(2(0(0(1(0(1(2(1(0(1(2(2(0(1(x0)))))))))))))))))
2(0(0(1(0(1(1(2(1(2(0(0(2(x0))))))))))))) 2(0(0(1(2(0(1(0(1(0(2(0(1(1(1(1(2(x0)))))))))))))))))
0(1(0(0(0(1(2(0(2(2(1(1(2(x0))))))))))))) 0(1(0(0(0(2(0(0(1(0(0(2(0(0(2(1(1(x0)))))))))))))))))
0(1(0(1(0(0(1(2(1(1(2(1(2(x0))))))))))))) 0(1(0(2(2(0(0(1(0(0(0(0(1(1(1(2(2(x0)))))))))))))))))
0(1(2(0(2(0(1(2(0(2(2(1(2(x0))))))))))))) 0(0(0(1(2(0(1(2(0(0(2(0(0(2(1(0(1(x0)))))))))))))))))
0(2(1(0(1(0(1(1(1(1(0(2(2(x0))))))))))))) 0(0(1(0(1(1(0(2(2(0(0(1(1(0(1(1(0(x0)))))))))))))))))

1.1 Bounds

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