Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96119)

The rewrite relation of the following TRS is considered.

0(1(2(x1))) 0(3(3(x1))) (1)
0(4(2(5(3(2(x1)))))) 0(1(2(4(0(x1))))) (2)
4(4(1(4(2(1(1(x1))))))) 5(0(1(1(1(5(x1)))))) (3)
4(5(0(3(2(5(3(x1))))))) 4(1(1(0(2(3(x1)))))) (4)
0(0(1(0(0(3(5(3(x1)))))))) 0(0(0(2(3(2(2(3(x1)))))))) (5)
5(1(1(3(1(5(4(2(x1)))))))) 5(5(3(4(4(2(0(2(x1)))))))) (6)
1(1(0(0(5(2(5(2(2(x1))))))))) 1(3(1(1(5(3(3(5(1(x1))))))))) (7)
4(3(2(4(0(0(2(2(3(5(0(x1))))))))))) 1(0(5(2(2(5(2(1(5(5(x1)))))))))) (8)
5(1(2(5(5(2(4(1(1(5(2(x1))))))))))) 5(1(5(2(0(3(5(0(1(4(1(x1))))))))))) (9)
1(0(4(3(2(1(4(1(5(1(1(2(x1)))))))))))) 4(2(0(3(3(5(3(1(4(5(3(3(5(x1))))))))))))) (10)
5(2(0(5(0(1(1(1(2(4(4(2(x1)))))))))))) 5(2(0(2(0(3(3(0(0(4(2(4(x1)))))))))))) (11)
4(1(5(5(5(3(3(1(3(3(3(4(5(x1))))))))))))) 4(1(4(3(4(5(1(1(1(1(2(5(x1)))))))))))) (12)
1(1(5(5(0(3(1(4(0(0(3(5(5(4(x1)))))))))))))) 0(1(3(1(3(0(3(4(5(0(4(1(5(4(x1)))))))))))))) (13)
3(1(3(0(1(4(5(2(5(0(2(2(4(3(x1)))))))))))))) 3(0(5(2(2(0(4(5(2(1(5(5(5(x1))))))))))))) (14)
5(3(1(5(4(5(4(5(1(1(1(2(2(1(5(2(x1)))))))))))))))) 5(0(0(5(2(3(2(4(4(1(1(1(5(5(5(x1))))))))))))))) (15)
0(3(0(2(1(3(4(2(1(5(2(4(3(0(1(0(2(x1))))))))))))))))) 2(0(4(5(5(2(0(0(1(0(3(2(5(4(1(3(3(x1))))))))))))))))) (16)
1(2(0(4(4(0(3(0(5(2(4(0(3(3(1(1(4(x1))))))))))))))))) 4(5(4(0(4(3(5(0(2(5(0(5(3(0(3(1(4(x1))))))))))))))))) (17)
4(0(1(5(4(1(3(5(5(0(0(5(4(5(4(0(5(4(x1)))))))))))))))))) 2(1(2(5(0(0(4(1(0(1(5(5(4(5(2(4(1(4(x1)))))))))))))))))) (18)
4(4(4(4(4(1(3(4(1(5(1(4(0(1(4(0(4(5(x1)))))))))))))))))) 4(2(2(3(1(2(3(5(4(3(3(4(2(1(2(0(3(4(5(x1))))))))))))))))))) (19)
4(3(3(3(5(5(1(2(5(0(4(1(4(4(2(3(0(4(2(x1))))))))))))))))))) 4(3(4(5(5(0(2(5(5(0(2(2(5(2(0(0(3(3(x1)))))))))))))))))) (20)
0(3(2(2(5(4(0(2(4(0(1(3(3(1(0(3(4(3(1(5(x1)))))))))))))))))))) 0(3(1(0(5(4(3(2(0(3(3(4(1(0(4(3(3(3(1(1(x1)))))))))))))))))))) (21)
1(0(5(2(2(3(2(0(5(5(3(4(4(3(1(0(0(2(1(2(x1)))))))))))))))))))) 1(3(1(4(5(3(3(3(3(0(2(5(1(4(5(4(3(1(5(1(x1)))))))))))))))))))) (22)
5(4(4(2(2(2(3(0(1(5(3(5(1(0(2(3(1(5(4(1(x1)))))))))))))))))))) 5(1(0(5(1(1(3(0(5(0(3(5(5(2(4(0(1(5(5(x1))))))))))))))))))) (23)
3(0(2(2(2(0(0(4(3(2(2(5(5(0(3(2(1(2(2(2(0(x1))))))))))))))))))))) 5(2(2(5(1(3(5(0(3(2(1(1(1(0(4(5(3(4(0(5(x1)))))))))))))))))))) (24)
4(4(1(2(0(1(2(3(1(0(1(0(0(3(3(4(4(1(1(1(2(x1))))))))))))))))))))) 4(1(0(1(0(3(1(4(1(5(3(3(3(0(1(2(3(1(2(5(1(x1))))))))))))))))))))) (25)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
2(1(0(x1))) 3(3(0(x1))) (26)
2(3(5(2(4(0(x1)))))) 0(4(2(1(0(x1))))) (27)
1(1(2(4(1(4(4(x1))))))) 5(1(1(1(0(5(x1)))))) (28)
3(5(2(3(0(5(4(x1))))))) 3(2(0(1(1(4(x1)))))) (29)
3(5(3(0(0(1(0(0(x1)))))))) 3(2(2(3(2(0(0(0(x1)))))))) (30)
2(4(5(1(3(1(1(5(x1)))))))) 2(0(2(4(4(3(5(5(x1)))))))) (31)
2(2(5(2(5(0(0(1(1(x1))))))))) 1(5(3(3(5(1(1(3(1(x1))))))))) (32)
0(5(3(2(2(0(0(4(2(3(4(x1))))))))))) 5(5(1(2(5(2(2(5(0(1(x1)))))))))) (33)
2(5(1(1(4(2(5(5(2(1(5(x1))))))))))) 1(4(1(0(5(3(0(2(5(1(5(x1))))))))))) (34)
2(1(1(5(1(4(1(2(3(4(0(1(x1)))))))))))) 5(3(3(5(4(1(3(5(3(3(0(2(4(x1))))))))))))) (35)
2(4(4(2(1(1(1(0(5(0(2(5(x1)))))))))))) 4(2(4(0(0(3(3(0(2(0(2(5(x1)))))))))))) (36)
5(4(3(3(3(1(3(3(5(5(5(1(4(x1))))))))))))) 5(2(1(1(1(1(5(4(3(4(1(4(x1)))))))))))) (37)
4(5(5(3(0(0(4(1(3(0(5(5(1(1(x1)))))))))))))) 4(5(1(4(0(5(4(3(0(3(1(3(1(0(x1)))))))))))))) (38)
3(4(2(2(0(5(2(5(4(1(0(3(1(3(x1)))))))))))))) 5(5(5(1(2(5(4(0(2(2(5(0(3(x1))))))))))))) (39)
2(5(1(2(2(1(1(1(5(4(5(4(5(1(3(5(x1)))))))))))))))) 5(5(5(1(1(1(4(4(2(3(2(5(0(0(5(x1))))))))))))))) (40)
2(0(1(0(3(4(2(5(1(2(4(3(1(2(0(3(0(x1))))))))))))))))) 3(3(1(4(5(2(3(0(1(0(0(2(5(5(4(0(2(x1))))))))))))))))) (41)
4(1(1(3(3(0(4(2(5(0(3(0(4(4(0(2(1(x1))))))))))))))))) 4(1(3(0(3(5(0(5(2(0(5(3(4(0(4(5(4(x1))))))))))))))))) (42)
4(5(0(4(5(4(5(0(0(5(5(3(1(4(5(1(0(4(x1)))))))))))))))))) 4(1(4(2(5(4(5(5(1(0(1(4(0(0(5(2(1(2(x1)))))))))))))))))) (43)
5(4(0(4(1(0(4(1(5(1(4(3(1(4(4(4(4(4(x1)))))))))))))))))) 5(4(3(0(2(1(2(4(3(3(4(5(3(2(1(3(2(2(4(x1))))))))))))))))))) (44)
2(4(0(3(2(4(4(1(4(0(5(2(1(5(5(3(3(3(4(x1))))))))))))))))))) 3(3(0(0(2(5(2(2(0(5(5(2(0(5(5(4(3(4(x1)))))))))))))))))) (45)
5(1(3(4(3(0(1(3(3(1(0(4(2(0(4(5(2(2(3(0(x1)))))))))))))))))))) 1(1(3(3(3(4(0(1(4(3(3(0(2(3(4(5(0(1(3(0(x1)))))))))))))))))))) (46)
2(1(2(0(0(1(3(4(4(3(5(5(0(2(3(2(2(5(0(1(x1)))))))))))))))))))) 1(5(1(3(4(5(4(1(5(2(0(3(3(3(3(5(4(1(3(1(x1)))))))))))))))))))) (47)
1(4(5(1(3(2(0(1(5(3(5(1(0(3(2(2(2(4(4(5(x1)))))))))))))))))))) 5(5(1(0(4(2(5(5(3(0(5(0(3(1(1(5(0(1(5(x1))))))))))))))))))) (48)
0(2(2(2(1(2(3(0(5(5(2(2(3(4(0(0(2(2(2(0(3(x1))))))))))))))))))))) 5(0(4(3(5(4(0(1(1(1(2(3(0(5(3(1(5(2(2(5(x1)))))))))))))))))))) (49)
2(1(1(1(4(4(3(3(0(0(1(0(1(3(2(1(0(2(1(4(4(x1))))))))))))))))))))) 1(5(2(1(3(2(1(0(3(3(3(5(1(4(1(3(0(1(0(1(4(x1))))))))))))))))))))) (50)

1.1 Bounds

The given TRS is match-(raise)-bounded by 2. This is shown by the following automaton. The automaton is closed under rewriting as it is compatible.