Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/157436)

The rewrite relation of the following TRS is considered.

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

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 492 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 2952 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 +
0
[51(x1)] = x1 +
34
[52(x1)] = x1 +
0
[53(x1)] = x1 +
0
[54(x1)] = x1 +
0
[55(x1)] = x1 +
0
[40(x1)] = x1 +
0
[41(x1)] = x1 +
347
[42(x1)] = x1 +
145
[43(x1)] = x1 +
351
[44(x1)] = x1 +
269
[45(x1)] = x1 +
0
[30(x1)] = x1 +
0
[31(x1)] = x1 +
34
[32(x1)] = x1 +
0
[33(x1)] = x1 +
0
[34(x1)] = x1 +
123
[35(x1)] = x1 +
92
[20(x1)] = x1 +
0
[21(x1)] = x1 +
79
[22(x1)] = x1 +
0
[23(x1)] = x1 +
0
[24(x1)] = x1 +
145
[25(x1)] = x1 +
0
[10(x1)] = x1 +
34
[11(x1)] = x1 +
113
[12(x1)] = x1 +
79
[13(x1)] = x1 +
81
[14(x1)] = x1 +
202
[15(x1)] = x1 +
353
[00(x1)] = x1 +
0
[01(x1)] = x1 +
158
[02(x1)] = x1 +
0
[03(x1)] = x1 +
164
[04(x1)] = x1 +
347
[05(x1)] = x1 +
347
all of the following rules can be deleted.

There are 2952 ruless (increase limit for explicit display).

1.1.1.1 R is empty

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