Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/4824)

The rewrite relation of the following TRS is considered.

5(5(x1)) 1(1(2(3(3(2(3(0(4(4(x1)))))))))) (1)
0(5(5(x1))) 1(0(1(3(3(0(1(2(4(4(x1)))))))))) (2)
5(2(5(x1))) 4(3(0(3(4(4(1(4(3(5(x1)))))))))) (3)
0(2(2(4(x1)))) 1(1(3(0(1(2(2(2(1(4(x1)))))))))) (4)
0(5(5(1(x1)))) 2(2(3(2(4(1(1(3(3(2(x1)))))))))) (5)
0(5(5(4(x1)))) 1(0(4(4(5(2(4(4(4(1(x1)))))))))) (6)
1(5(4(0(x1)))) 0(4(4(2(3(3(0(3(3(2(x1)))))))))) (7)
5(4(0(0(x1)))) 0(4(3(0(4(3(3(5(0(0(x1)))))))))) (8)
5(5(3(4(x1)))) 1(3(2(3(3(0(0(3(2(4(x1)))))))))) (9)
0(5(0(2(1(x1))))) 1(2(1(2(2(3(0(1(2(0(x1)))))))))) (10)
2(2(5(3(5(x1))))) 0(1(1(5(3(0(3(3(1(4(x1)))))))))) (11)
2(4(0(5(0(x1))))) 1(1(2(4(4(1(2(2(5(0(x1)))))))))) (12)
2(4(0(5(4(x1))))) 2(4(3(1(4(4(0(5(4(4(x1)))))))))) (13)
3(5(3(4(0(x1))))) 3(0(3(3(5(2(3(3(4(0(x1)))))))))) (14)
0(0(5(1(1(5(x1)))))) 3(1(0(3(3(0(3(2(1(5(x1)))))))))) (15)
0(5(5(0(5(5(x1)))))) 2(0(2(2(1(1(0(4(0(5(x1)))))))))) (16)
0(5(5(1(1(2(x1)))))) 2(4(3(1(2(1(2(0(4(2(x1)))))))))) (17)
1(5(4(0(5(5(x1)))))) 3(0(3(0(4(5(1(5(1(5(x1)))))))))) (18)
1(5(5(5(0(5(x1)))))) 0(2(0(3(2(0(3(2(4(5(x1)))))))))) (19)
2(2(0(2(3(4(x1)))))) 3(3(2(3(2(1(2(3(2(4(x1)))))))))) (20)
2(5(4(0(2(5(x1)))))) 3(2(0(4(0(2(3(0(4(5(x1)))))))))) (21)
3(4(2(5(0(0(x1)))))) 0(4(0(0(3(0(3(3(0(0(x1)))))))))) (22)
4(0(0(0(4(0(x1)))))) 1(4(3(2(4(3(5(0(2(0(x1)))))))))) (23)
4(0(0(5(3(4(x1)))))) 4(0(3(1(1(1(0(0(0(4(x1)))))))))) (24)
4(0(5(2(4(0(x1)))))) 2(1(3(3(0(4(2(2(0(0(x1)))))))))) (25)
4(0(5(2(5(0(x1)))))) 3(2(1(2(3(5(0(4(1(1(x1)))))))))) (26)
4(3(0(2(5(0(x1)))))) 4(1(3(4(1(1(0(1(5(0(x1)))))))))) (27)
5(1(4(0(5(2(x1)))))) 5(4(5(1(5(1(2(0(3(3(x1)))))))))) (28)
5(1(5(0(2(5(x1)))))) 5(2(2(5(3(1(4(1(4(4(x1)))))))))) (29)
5(2(5(2(5(2(x1)))))) 1(1(4(4(5(4(0(4(4(2(x1)))))))))) (30)
5(3(5(5(3(2(x1)))))) 3(0(4(5(4(1(5(0(3(2(x1)))))))))) (31)
5(5(1(2(5(2(x1)))))) 5(1(3(4(4(0(3(2(4(1(x1)))))))))) (32)
5(5(3(4(5(1(x1)))))) 2(1(2(0(3(3(1(0(0(5(x1)))))))))) (33)
0(5(3(2(5(1(0(x1))))))) 3(3(0(1(3(4(5(1(4(0(x1)))))))))) (34)
2(2(5(3(4(0(1(x1))))))) 2(1(2(2(4(0(2(0(1(3(x1)))))))))) (35)
2(3(2(5(1(0(5(x1))))))) 2(0(3(3(3(1(0(5(4(5(x1)))))))))) (36)
2(4(1(0(5(5(5(x1))))))) 3(3(3(2(2(1(4(4(5(5(x1)))))))))) (37)
2(4(3(1(5(5(3(x1))))))) 0(1(0(4(2(2(2(0(0(3(x1)))))))))) (38)
4(0(2(4(5(2(5(x1))))))) 2(0(2(0(4(4(1(3(5(5(x1)))))))))) (39)
4(0(5(2(4(2(5(x1))))))) 4(4(4(2(4(2(0(2(1(5(x1)))))))))) (40)
4(1(2(4(5(1(4(x1))))))) 4(4(3(3(4(0(4(4(5(4(x1)))))))))) (41)
5(1(1(5(2(0(0(x1))))))) 2(3(0(1(1(1(2(3(1(0(x1)))))))))) (42)
5(2(2(5(3(0(2(x1))))))) 1(4(2(1(0(2(3(2(4(4(x1)))))))))) (43)
5(3(4(0(1(5(3(x1))))))) 4(4(4(2(3(0(4(3(5(3(x1)))))))))) (44)
5(4(2(3(1(5(0(x1))))))) 1(2(2(3(3(0(0(2(4(1(x1)))))))))) (45)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

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.