Certification Problem

Input (TPDB SRS_Relative/Zantema_06_relative/rel09)

The relative rewrite relation R/S is considered where R is the following TRS

b(q(b(x1))) → b(p(b(x1))) (1)

and S is the following TRS.

0(p(0(x1))) → q(x1) (2)
1(p(1(x1))) → q(x1) (3)
0(q(0(x1))) → q(x1) (4)
1(q(1(x1))) → q(x1) (5)
p(x1) → 1(p(1(0(1(x1))))) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Split

We split R in the relative problem D/R-D and R-D, where the rules D

0(p(0(x1))) → q(x1) (2)
are deleted.

1.1 Closure Under Flat Contexts

Using the flat contexts

{1(☐), 0(☐), b(☐), q(☐), p(☐)}

We obtain the transformed TRS
1(0(p(0(x1)))) → 1(q(x1)) (7)
0(0(p(0(x1)))) → 0(q(x1)) (8)
b(0(p(0(x1)))) → b(q(x1)) (9)
q(0(p(0(x1)))) → q(q(x1)) (10)
p(0(p(0(x1)))) → p(q(x1)) (11)
1(b(q(b(x1)))) → 1(b(p(b(x1)))) (12)
1(1(p(1(x1)))) → 1(q(x1)) (13)
1(0(q(0(x1)))) → 1(q(x1)) (14)
1(1(q(1(x1)))) → 1(q(x1)) (15)
1(p(x1)) → 1(1(p(1(0(1(x1)))))) (16)
0(b(q(b(x1)))) → 0(b(p(b(x1)))) (17)
0(1(p(1(x1)))) → 0(q(x1)) (18)
0(0(q(0(x1)))) → 0(q(x1)) (19)
0(1(q(1(x1)))) → 0(q(x1)) (20)
0(p(x1)) → 0(1(p(1(0(1(x1)))))) (21)
b(b(q(b(x1)))) → b(b(p(b(x1)))) (22)
b(1(p(1(x1)))) → b(q(x1)) (23)
b(0(q(0(x1)))) → b(q(x1)) (24)
b(1(q(1(x1)))) → b(q(x1)) (25)
b(p(x1)) → b(1(p(1(0(1(x1)))))) (26)
q(b(q(b(x1)))) → q(b(p(b(x1)))) (27)
q(1(p(1(x1)))) → q(q(x1)) (28)
q(0(q(0(x1)))) → q(q(x1)) (29)
q(1(q(1(x1)))) → q(q(x1)) (30)
q(p(x1)) → q(1(p(1(0(1(x1)))))) (31)
p(b(q(b(x1)))) → p(b(p(b(x1)))) (32)
p(1(p(1(x1)))) → p(q(x1)) (33)
p(0(q(0(x1)))) → p(q(x1)) (34)
p(1(q(1(x1)))) → p(q(x1)) (35)
p(p(x1)) → p(1(p(1(0(1(x1)))))) (36)

1.1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,...,4}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 5):

[1(x1)] = 5x1 + 0
[0(x1)] = 5x1 + 1
[b(x1)] = 5x1 + 2
[q(x1)] = 5x1 + 3
[p(x1)] = 5x1 + 4

We obtain the labeled TRS

There are 150 ruless (increase limit for explicit display).

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[10(x1)] = x1 +
0
[11(x1)] = x1 +
0
[12(x1)] = x1 +
0
[13(x1)] = x1 +
1
[14(x1)] = x1 +
1
[00(x1)] = x1 +
0
[01(x1)] = x1 +
3
[02(x1)] = x1 +
3
[03(x1)] = x1 +
1
[04(x1)] = x1 +
3
[b0(x1)] = x1 +
0
[b1(x1)] = x1 +
3
[b2(x1)] = x1 +
0
[b3(x1)] = x1 +
1
[b4(x1)] = x1 +
1
[q0(x1)] = x1 +
0
[q1(x1)] = x1 +
0
[q2(x1)] = x1 +
0
[q3(x1)] = x1 +
0
[q4(x1)] = x1 +
1
[p0(x1)] = x1 +
0
[p1(x1)] = x1 +
3
[p2(x1)] = x1 +
0
[p3(x1)] = x1 +
1
[p4(x1)] = x1 +
3
all of the following rules can be deleted.
b1(04(p1(02(x1)))) → b3(q2(x1)) (37)
b1(04(p1(03(x1)))) → b3(q3(x1)) (38)
b1(04(p1(04(x1)))) → b3(q4(x1)) (39)
b1(04(p1(01(x1)))) → b3(q1(x1)) (40)
b1(04(p1(00(x1)))) → b3(q0(x1)) (41)
q1(04(p1(02(x1)))) → q3(q2(x1)) (42)
q1(04(p1(03(x1)))) → q3(q3(x1)) (43)
q1(04(p1(04(x1)))) → q3(q4(x1)) (44)
q1(04(p1(01(x1)))) → q3(q1(x1)) (45)
q1(04(p1(00(x1)))) → q3(q0(x1)) (46)
p1(04(p1(02(x1)))) → p3(q2(x1)) (47)
p1(04(p1(03(x1)))) → p3(q3(x1)) (48)
p1(04(p1(04(x1)))) → p3(q4(x1)) (49)
p1(04(p1(01(x1)))) → p3(q1(x1)) (50)
p1(04(p1(00(x1)))) → p3(q0(x1)) (51)
01(04(p1(02(x1)))) → 03(q2(x1)) (52)
01(04(p1(03(x1)))) → 03(q3(x1)) (53)
01(04(p1(04(x1)))) → 03(q4(x1)) (54)
01(04(p1(01(x1)))) → 03(q1(x1)) (55)
01(04(p1(00(x1)))) → 03(q0(x1)) (56)
11(04(p1(02(x1)))) → 13(q2(x1)) (57)
11(04(p1(03(x1)))) → 13(q3(x1)) (58)
11(04(p1(04(x1)))) → 13(q4(x1)) (59)
11(04(p1(01(x1)))) → 13(q1(x1)) (60)
11(04(p1(00(x1)))) → 13(q0(x1)) (61)
b0(14(p0(13(x1)))) → b3(q3(x1)) (88)
q0(14(p0(12(x1)))) → q3(q2(x1)) (92)
q0(14(p0(13(x1)))) → q3(q3(x1)) (93)
q0(14(p0(14(x1)))) → q3(q4(x1)) (94)
q0(14(p0(11(x1)))) → q3(q1(x1)) (95)
q0(14(p0(10(x1)))) → q3(q0(x1)) (96)
p0(14(p0(13(x1)))) → p3(q3(x1)) (98)
00(14(p0(13(x1)))) → 03(q3(x1)) (103)
10(14(p0(13(x1)))) → 13(q3(x1)) (108)
b1(03(q1(02(x1)))) → b3(q2(x1)) (112)
b1(03(q1(03(x1)))) → b3(q3(x1)) (113)
b1(03(q1(04(x1)))) → b3(q4(x1)) (114)
b1(03(q1(01(x1)))) → b3(q1(x1)) (115)
b1(03(q1(00(x1)))) → b3(q0(x1)) (116)
q1(03(q1(02(x1)))) → q3(q2(x1)) (117)
q1(03(q1(03(x1)))) → q3(q3(x1)) (118)
q1(03(q1(04(x1)))) → q3(q4(x1)) (119)
q1(03(q1(01(x1)))) → q3(q1(x1)) (120)
q1(03(q1(00(x1)))) → q3(q0(x1)) (121)
p1(03(q1(02(x1)))) → p3(q2(x1)) (122)
p1(03(q1(03(x1)))) → p3(q3(x1)) (123)
p1(03(q1(04(x1)))) → p3(q4(x1)) (124)
p1(03(q1(01(x1)))) → p3(q1(x1)) (125)
p1(03(q1(00(x1)))) → p3(q0(x1)) (126)
01(03(q1(02(x1)))) → 03(q2(x1)) (127)
01(03(q1(03(x1)))) → 03(q3(x1)) (128)
01(03(q1(04(x1)))) → 03(q4(x1)) (129)
01(03(q1(01(x1)))) → 03(q1(x1)) (130)
01(03(q1(00(x1)))) → 03(q0(x1)) (131)
11(03(q1(02(x1)))) → 13(q2(x1)) (132)
11(03(q1(03(x1)))) → 13(q3(x1)) (133)
11(03(q1(04(x1)))) → 13(q4(x1)) (134)
11(03(q1(01(x1)))) → 13(q1(x1)) (135)
b0(13(q0(13(x1)))) → b3(q3(x1)) (138)
q0(13(q0(12(x1)))) → q3(q2(x1)) (142)
q0(13(q0(13(x1)))) → q3(q3(x1)) (143)
q0(13(q0(14(x1)))) → q3(q4(x1)) (144)
q0(13(q0(11(x1)))) → q3(q1(x1)) (145)
q0(13(q0(10(x1)))) → q3(q0(x1)) (146)
p0(13(q0(13(x1)))) → p3(q3(x1)) (148)
00(13(q0(13(x1)))) → 03(q3(x1)) (153)
10(13(q0(13(x1)))) → 13(q3(x1)) (158)
b4(p4(x1)) → b0(14(p0(11(00(14(x1)))))) (164)
b4(p1(x1)) → b0(14(p0(11(00(11(x1)))))) (165)
q4(p4(x1)) → q0(14(p0(11(00(14(x1)))))) (169)
q4(p1(x1)) → q0(14(p0(11(00(11(x1)))))) (170)
p4(p2(x1)) → p0(14(p0(11(00(12(x1)))))) (172)
p4(p3(x1)) → p0(14(p0(11(00(13(x1)))))) (173)
p4(p4(x1)) → p0(14(p0(11(00(14(x1)))))) (174)
p4(p1(x1)) → p0(14(p0(11(00(11(x1)))))) (175)
p4(p0(x1)) → p0(14(p0(11(00(10(x1)))))) (176)
04(p2(x1)) → 00(14(p0(11(00(12(x1)))))) (177)
04(p3(x1)) → 00(14(p0(11(00(13(x1)))))) (178)
04(p4(x1)) → 00(14(p0(11(00(14(x1)))))) (179)
04(p1(x1)) → 00(14(p0(11(00(11(x1)))))) (180)
04(p0(x1)) → 00(14(p0(11(00(10(x1)))))) (181)
14(p4(x1)) → 10(14(p0(11(00(14(x1)))))) (184)
14(p1(x1)) → 10(14(p0(11(00(11(x1)))))) (185)

1.1.1.1.1 R is empty

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

1.2 Rule Removal

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
[1(x1)] =
0 -5 1
-∞ 0 -∞
-∞ -6 0
· x1
[0(x1)] =
0 -6 3
-3 -∞ 0
-3 -1 0
· x1
[b(x1)] =
1 -∞ -∞
1 0 1
0 -∞ 0
· x1
[q(x1)] =
2 5 -∞
-∞ -∞ 4
1 2 3
· x1
[p(x1)] =
3 5 6
3 5 6
2 5 5
· x1
all of the following rules can be deleted.
1(p(1(x1))) → q(x1) (3)

1.2.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[1(x1)] = x1 +
0
[0(x1)] = x1 +
0
[b(x1)] = x1 +
0
[q(x1)] = x1 +
1
[p(x1)] = x1 +
0
all of the following rules can be deleted.
b(q(b(x1))) → b(p(b(x1))) (1)

1.2.1.1 R is empty

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