Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/41838)

The rewrite relation of the following TRS is considered.

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

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 468 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 2808 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 +
1
[52(x1)] = x1 +
0
[53(x1)] = x1 +
0
[54(x1)] = x1 +
0
[55(x1)] = x1 +
0
[40(x1)] = x1 +
1
[41(x1)] = x1 +
1
[42(x1)] = x1 +
3
[43(x1)] = x1 +
0
[44(x1)] = x1 +
0
[45(x1)] = x1 +
3
[30(x1)] = x1 +
1
[31(x1)] = x1 +
0
[32(x1)] = x1 +
1
[33(x1)] = x1 +
0
[34(x1)] = x1 +
0
[35(x1)] = x1 +
1
[20(x1)] = x1 +
0
[21(x1)] = x1 +
0
[22(x1)] = x1 +
1
[23(x1)] = x1 +
0
[24(x1)] = x1 +
0
[25(x1)] = x1 +
0
[10(x1)] = x1 +
1
[11(x1)] = x1 +
0
[12(x1)] = x1 +
0
[13(x1)] = x1 +
0
[14(x1)] = x1 +
1
[15(x1)] = x1 +
0
[00(x1)] = x1 +
8
[01(x1)] = x1 +
3
[02(x1)] = x1 +
8
[03(x1)] = x1 +
0
[04(x1)] = x1 +
11
[05(x1)] = x1 +
0
all of the following rules can be deleted.

There are 2808 ruless (increase limit for explicit display).

1.1.1.1 R is empty

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