Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/57355)

The rewrite relation of the following TRS is considered.

0(1(0(2(x1)))) → 2(0(3(1(0(x1))))) (1)
0(1(0(2(x1)))) → 2(0(0(3(1(2(x1)))))) (2)
0(1(0(2(x1)))) → 2(0(3(1(0(4(x1)))))) (3)
0(1(0(2(x1)))) → 2(2(0(3(1(0(x1)))))) (4)
0(1(0(2(x1)))) → 2(3(1(0(0(2(x1)))))) (5)
0(1(0(2(x1)))) → 2(3(1(0(3(0(x1)))))) (6)
0(1(0(2(x1)))) → 4(1(0(3(0(2(x1)))))) (7)
0(1(0(2(x1)))) → 4(1(0(4(0(2(x1)))))) (8)
0(1(4(2(x1)))) → 2(3(1(0(4(x1))))) (9)
0(1(4(2(x1)))) → 2(4(0(3(1(x1))))) (10)
0(1(4(2(x1)))) → 3(2(1(0(4(x1))))) (11)
0(1(4(2(x1)))) → 3(2(1(4(0(x1))))) (12)
0(1(4(2(x1)))) → 4(0(3(1(2(x1))))) (13)
0(1(4(2(x1)))) → 4(1(0(3(2(x1))))) (14)
0(1(4(2(x1)))) → 4(1(0(4(2(x1))))) (15)
0(1(4(2(x1)))) → 4(1(0(5(2(x1))))) (16)
0(1(4(2(x1)))) → 2(0(3(1(0(4(x1)))))) (17)
0(1(4(2(x1)))) → 2(0(3(1(4(4(x1)))))) (18)
0(1(4(2(x1)))) → 2(3(1(4(0(4(x1)))))) (19)
0(1(4(2(x1)))) → 2(4(3(0(4(1(x1)))))) (20)
0(1(4(2(x1)))) → 2(4(3(1(0(3(x1)))))) (21)
0(1(4(2(x1)))) → 3(2(1(0(4(1(x1)))))) (22)
0(1(4(2(x1)))) → 3(2(2(1(4(0(x1)))))) (23)
0(1(4(2(x1)))) → 3(2(3(1(0(4(x1)))))) (24)
0(1(4(2(x1)))) → 3(2(3(1(4(0(x1)))))) (25)
0(1(4(2(x1)))) → 4(0(3(1(3(2(x1)))))) (26)
0(1(4(2(x1)))) → 4(0(3(1(4(2(x1)))))) (27)
0(1(4(2(x1)))) → 4(1(0(4(3(2(x1)))))) (28)
0(1(4(2(x1)))) → 4(1(0(4(5(2(x1)))))) (29)
0(1(4(2(x1)))) → 4(1(0(5(3(2(x1)))))) (30)
0(1(4(2(x1)))) → 4(1(1(0(5(2(x1)))))) (31)
0(1(4(2(x1)))) → 4(1(3(0(5(2(x1)))))) (32)
0(1(4(2(x1)))) → 4(3(0(3(1(2(x1)))))) (33)
0(1(4(2(x1)))) → 4(4(0(3(1(2(x1)))))) (34)
0(0(1(0(2(x1))))) → 1(0(0(2(0(4(x1)))))) (35)
0(0(1(0(2(x1))))) → 1(0(4(0(0(2(x1)))))) (36)
0(0(1(0(2(x1))))) → 2(1(0(3(0(0(x1)))))) (37)
0(0(1(4(2(x1))))) → 0(0(3(1(2(4(x1)))))) (38)
0(0(1(4(2(x1))))) → 0(2(3(1(0(4(x1)))))) (39)
0(0(1(4(2(x1))))) → 0(2(4(0(3(1(x1)))))) (40)
0(0(1(4(2(x1))))) → 0(3(1(0(2(4(x1)))))) (41)
0(0(1(4(2(x1))))) → 1(0(3(4(0(2(x1)))))) (42)
0(0(1(4(2(x1))))) → 2(0(0(3(1(4(x1)))))) (43)
0(0(1(4(2(x1))))) → 2(1(0(4(0(0(x1)))))) (44)
0(1(2(0(2(x1))))) → 2(0(1(0(4(2(x1)))))) (45)
0(1(2(4(2(x1))))) → 2(3(1(0(2(4(x1)))))) (46)
0(1(2(4(2(x1))))) → 4(1(0(2(2(4(x1)))))) (47)
0(1(3(4(2(x1))))) → 2(3(1(4(4(0(x1)))))) (48)
0(1(3(4(2(x1))))) → 2(4(3(0(4(1(x1)))))) (49)
0(1(3(4(2(x1))))) → 3(2(1(0(4(0(x1)))))) (50)
0(1(3(4(2(x1))))) → 4(0(3(3(1(2(x1)))))) (51)
0(1(3(4(2(x1))))) → 4(1(4(0(3(2(x1)))))) (52)
0(1(4(0(2(x1))))) → 2(0(3(1(0(4(x1)))))) (53)
0(1(5(0(2(x1))))) → 0(2(3(1(0(5(x1)))))) (54)
0(1(5(0(2(x1))))) → 3(0(5(1(0(2(x1)))))) (55)
0(1(5(0(2(x1))))) → 5(1(3(0(0(2(x1)))))) (56)
0(1(5(4(2(x1))))) → 0(4(4(1(2(5(x1)))))) (57)
0(1(5(4(2(x1))))) → 1(0(4(5(1(2(x1)))))) (58)
0(1(5(4(2(x1))))) → 2(0(4(4(5(1(x1)))))) (59)
0(1(5(4(2(x1))))) → 4(0(2(3(1(5(x1)))))) (60)
0(1(5(4(2(x1))))) → 4(1(0(2(5(2(x1)))))) (61)
0(1(5(4(2(x1))))) → 4(1(0(5(2(5(x1)))))) (62)
0(1(5(4(2(x1))))) → 4(2(1(3(0(5(x1)))))) (63)
0(1(5(4(2(x1))))) → 4(3(1(0(2(5(x1)))))) (64)
0(1(5(4(2(x1))))) → 4(4(0(5(1(2(x1)))))) (65)
0(1(5(4(2(x1))))) → 4(4(2(1(0(5(x1)))))) (66)
0(1(5(4(2(x1))))) → 5(0(4(5(2(1(x1)))))) (67)
0(1(5(4(2(x1))))) → 5(1(2(0(4(3(x1)))))) (68)
0(1(5(4(2(x1))))) → 5(3(1(0(4(2(x1)))))) (69)
0(2(1(4(2(x1))))) → 0(4(4(1(2(2(x1)))))) (70)
0(2(1(4(2(x1))))) → 3(2(2(1(4(0(x1)))))) (71)
0(2(1(4(2(x1))))) → 4(1(0(3(2(2(x1)))))) (72)
5(0(1(4(2(x1))))) → 2(0(4(3(5(1(x1)))))) (73)
5(0(1(4(2(x1))))) → 2(4(0(3(1(5(x1)))))) (74)
5(0(1(4(2(x1))))) → 4(1(0(5(3(2(x1)))))) (75)
5(0(1(4(2(x1))))) → 5(2(1(1(0(4(x1)))))) (76)
5(0(2(0(2(x1))))) → 5(0(3(0(2(2(x1)))))) (77)
5(0(2(0(2(x1))))) → 5(0(4(0(2(2(x1)))))) (78)
5(0(2(4(2(x1))))) → 5(4(0(3(2(2(x1)))))) (79)
5(1(5(0(2(x1))))) → 5(2(1(4(5(0(x1)))))) (80)
5(1(5(4(2(x1))))) → 5(2(1(0(4(5(x1)))))) (81)
5(4(1(4(2(x1))))) → 3(2(1(4(4(5(x1)))))) (82)
5(4(1(4(2(x1))))) → 4(4(3(5(1(2(x1)))))) (83)
5(4(2(0(2(x1))))) → 3(0(5(2(2(4(x1)))))) (84)
5(4(2(0(2(x1))))) → 4(0(5(3(2(2(x1)))))) (85)
5(4(2(0(2(x1))))) → 5(2(2(2(4(0(x1)))))) (86)
5(4(2(0(2(x1))))) → 5(3(2(2(4(0(x1)))))) (87)
5(4(2(0(2(x1))))) → 5(4(2(2(4(0(x1)))))) (88)
5(4(2(4(2(x1))))) → 0(4(4(5(2(2(x1)))))) (89)
5(4(2(4(2(x1))))) → 5(4(4(3(2(2(x1)))))) (90)
5(4(5(4(2(x1))))) → 4(5(0(4(5(2(x1)))))) (91)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{5(☐), 4(☐), 3(☐), 2(☐), 1(☐), 0(☐)}

We obtain the transformed TRS

There are 546 ruless (increase limit for explicit display).

1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,...,5}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 6):

[5(x1)] = 6x1 + 0
[4(x1)] = 6x1 + 1
[3(x1)] = 6x1 + 2
[2(x1)] = 6x1 + 3
[1(x1)] = 6x1 + 4
[0(x1)] = 6x1 + 5

We obtain the labeled TRS

There are 3276 ruless (increase limit for explicit display).

1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[50(x1)] = x1 +
12
[51(x1)] = x1 +
48
[52(x1)] = x1 +
0
[53(x1)] = x1 +
0
[54(x1)] = x1 +
12
[55(x1)] = x1 +
36
[40(x1)] = x1 +
0
[41(x1)] = x1 +
0
[42(x1)] = x1 +
0
[43(x1)] = x1 +
37
[44(x1)] = x1 +
0
[45(x1)] = x1 +
2
[30(x1)] = x1 +
12
[31(x1)] = x1 +
36
[32(x1)] = x1 +
0
[33(x1)] = x1 +
0
[34(x1)] = x1 +
0
[35(x1)] = x1 +
2
[20(x1)] = x1 +
36
[21(x1)] = x1 +
7
[22(x1)] = x1 +
0
[23(x1)] = x1 +
36
[24(x1)] = x1 +
43
[25(x1)] = x1 +
12
[10(x1)] = x1 +
48
[11(x1)] = x1 +
38
[12(x1)] = x1 +
0
[13(x1)] = x1 +
43
[14(x1)] = x1 +
0
[15(x1)] = x1 +
11
[00(x1)] = x1 +
1
[01(x1)] = x1 +
0
[02(x1)] = x1 +
0
[03(x1)] = x1 +
43
[04(x1)] = x1 +
52
[05(x1)] = x1 +
14
all of the following rules can be deleted.

There are 3276 ruless (increase limit for explicit display).

1.1.1.1 R is empty

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