Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/68989)

The rewrite relation of the following TRS is considered.

0(0(1(2(1(x1))))) → 1(0(1(1(0(x1))))) (1)
0(3(1(2(4(x1))))) → 0(5(1(4(x1)))) (2)
4(3(0(1(1(x1))))) → 5(4(5(2(x1)))) (3)
4(3(4(2(2(x1))))) → 0(5(0(2(x1)))) (4)
2(1(1(4(0(2(x1)))))) → 1(1(5(3(5(x1))))) (5)
2(2(4(1(3(4(2(x1))))))) → 2(3(2(4(3(5(x1)))))) (6)
4(0(5(4(2(4(0(x1))))))) → 0(1(2(2(2(2(0(x1))))))) (7)
4(2(4(2(5(1(0(1(5(x1))))))))) → 5(0(3(1(0(2(2(5(x1)))))))) (8)
5(2(4(5(0(3(0(2(3(x1))))))))) → 5(3(2(1(0(1(3(4(0(x1))))))))) (9)
1(4(1(3(2(3(3(1(2(1(x1)))))))))) → 1(5(3(2(4(5(1(3(4(x1))))))))) (10)
4(0(4(2(3(4(5(1(1(5(1(x1))))))))))) → 5(4(3(2(2(3(2(1(2(4(0(x1))))))))))) (11)
4(5(2(2(5(4(4(3(4(5(4(x1))))))))))) → 5(2(3(2(5(0(0(0(5(4(x1)))))))))) (12)
4(3(4(4(0(3(0(3(2(3(2(1(x1)))))))))))) → 5(5(5(5(5(0(2(2(4(4(2(0(x1)))))))))))) (13)
5(3(4(4(3(3(5(2(5(2(1(1(4(2(x1)))))))))))))) → 5(4(0(5(2(5(5(3(1(0(3(3(5(x1))))))))))))) (14)
0(5(1(0(3(3(2(5(5(4(0(5(5(5(2(x1))))))))))))))) → 1(0(1(3(4(5(3(3(3(2(2(3(3(5(2(x1))))))))))))))) (15)
4(5(3(2(1(1(5(2(2(3(4(3(2(3(1(x1))))))))))))))) → 0(1(3(5(0(1(3(4(0(3(5(4(3(1(x1)))))))))))))) (16)
5(0(1(0(1(1(5(1(1(5(5(2(1(1(0(x1))))))))))))))) → 5(1(5(1(1(1(3(0(3(3(3(3(1(0(x1)))))))))))))) (17)
5(3(0(4(4(1(1(5(3(4(1(1(2(3(2(x1))))))))))))))) → 5(4(1(4(0(2(1(2(2(5(3(5(3(4(4(x1))))))))))))))) (18)
2(4(1(0(2(3(2(3(5(3(1(2(3(1(1(4(x1)))))))))))))))) → 2(2(2(1(4(5(0(1(0(3(1(3(5(1(2(x1))))))))))))))) (19)
0(1(1(3(2(2(0(0(0(5(0(2(4(3(3(0(1(x1))))))))))))))))) → 5(4(1(1(0(5(2(0(2(3(3(3(0(5(0(1(x1)))))))))))))))) (20)
2(1(1(5(3(1(3(4(3(5(3(3(2(4(3(1(4(x1))))))))))))))))) → 1(0(1(0(0(2(1(3(2(2(0(3(0(5(2(4(x1)))))))))))))))) (21)
2(4(3(5(0(2(5(5(1(5(0(4(4(4(1(4(3(x1))))))))))))))))) → 2(0(5(2(2(0(5(4(1(3(2(4(1(4(1(1(0(x1))))))))))))))))) (22)
0(4(5(4(5(0(2(3(1(2(4(5(3(5(0(4(3(3(2(x1))))))))))))))))))) → 1(0(2(4(5(5(2(2(4(2(1(1(4(0(1(2(3(0(2(5(x1)))))))))))))))))))) (23)
4(5(1(0(2(0(5(4(5(4(4(2(5(5(2(3(5(4(2(3(x1)))))))))))))))))))) → 0(5(5(2(0(5(2(4(2(5(2(5(2(0(1(5(2(3(3(0(x1)))))))))))))))))))) (24)
3(1(2(4(3(4(3(2(0(3(2(3(4(3(4(5(4(3(4(1(1(x1))))))))))))))))))))) → 3(4(0(0(2(4(5(0(0(4(3(5(4(3(0(3(2(2(1(1(x1)))))))))))))))))))) (25)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[5(x1)] = x1 +
37
[4(x1)] = x1 +
36
[3(x1)] = x1 +
41
[2(x1)] = x1 +
32
[1(x1)] = x1 +
32
[0(x1)] = x1 +
40
all of the following rules can be deleted.
0(3(1(2(4(x1))))) → 0(5(1(4(x1)))) (2)
4(3(0(1(1(x1))))) → 5(4(5(2(x1)))) (3)
4(3(4(2(2(x1))))) → 0(5(0(2(x1)))) (4)
2(1(1(4(0(2(x1)))))) → 1(1(5(3(5(x1))))) (5)
2(2(4(1(3(4(2(x1))))))) → 2(3(2(4(3(5(x1)))))) (6)
4(0(5(4(2(4(0(x1))))))) → 0(1(2(2(2(2(0(x1))))))) (7)
4(2(4(2(5(1(0(1(5(x1))))))))) → 5(0(3(1(0(2(2(5(x1)))))))) (8)
5(2(4(5(0(3(0(2(3(x1))))))))) → 5(3(2(1(0(1(3(4(0(x1))))))))) (9)
1(4(1(3(2(3(3(1(2(1(x1)))))))))) → 1(5(3(2(4(5(1(3(4(x1))))))))) (10)
4(5(2(2(5(4(4(3(4(5(4(x1))))))))))) → 5(2(3(2(5(0(0(0(5(4(x1)))))))))) (12)
4(3(4(4(0(3(0(3(2(3(2(1(x1)))))))))))) → 5(5(5(5(5(0(2(2(4(4(2(0(x1)))))))))))) (13)
5(3(4(4(3(3(5(2(5(2(1(1(4(2(x1)))))))))))))) → 5(4(0(5(2(5(5(3(1(0(3(3(5(x1))))))))))))) (14)
4(5(3(2(1(1(5(2(2(3(4(3(2(3(1(x1))))))))))))))) → 0(1(3(5(0(1(3(4(0(3(5(4(3(1(x1)))))))))))))) (16)
5(0(1(0(1(1(5(1(1(5(5(2(1(1(0(x1))))))))))))))) → 5(1(5(1(1(1(3(0(3(3(3(3(1(0(x1)))))))))))))) (17)
2(4(1(0(2(3(2(3(5(3(1(2(3(1(1(4(x1)))))))))))))))) → 2(2(2(1(4(5(0(1(0(3(1(3(5(1(2(x1))))))))))))))) (19)
0(1(1(3(2(2(0(0(0(5(0(2(4(3(3(0(1(x1))))))))))))))))) → 5(4(1(1(0(5(2(0(2(3(3(3(0(5(0(1(x1)))))))))))))))) (20)
2(1(1(5(3(1(3(4(3(5(3(3(2(4(3(1(4(x1))))))))))))))))) → 1(0(1(0(0(2(1(3(2(2(0(3(0(5(2(4(x1)))))))))))))))) (21)
2(4(3(5(0(2(5(5(1(5(0(4(4(4(1(4(3(x1))))))))))))))))) → 2(0(5(2(2(0(5(4(1(3(2(4(1(4(1(1(0(x1))))))))))))))))) (22)
0(4(5(4(5(0(2(3(1(2(4(5(3(5(0(4(3(3(2(x1))))))))))))))))))) → 1(0(2(4(5(5(2(2(4(2(1(1(4(0(1(2(3(0(2(5(x1)))))))))))))))))))) (23)
3(1(2(4(3(4(3(2(0(3(2(3(4(3(4(5(4(3(4(1(1(x1))))))))))))))))))))) → 3(4(0(0(2(4(5(0(0(4(3(5(4(3(0(3(2(2(1(1(x1)))))))))))))))))))) (25)

1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
1(2(1(0(0(x1))))) → 0(1(1(0(1(x1))))) (26)
1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 0(4(2(1(2(3(2(2(3(4(5(x1))))))))))) (27)
2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 2(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) (28)
2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 4(4(3(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))))) (29)
3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 0(3(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))))) (30)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 3#(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))))))) (31)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 3#(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))) (32)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 2#(5(5(0(x1)))) (33)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 2#(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))) (34)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 2#(5(2(4(2(5(0(2(5(5(0(x1))))))))))) (35)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 2#(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))))) (36)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 2#(5(0(2(5(5(0(x1))))))) (37)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 2#(4(2(5(0(2(5(5(0(x1))))))))) (38)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 1#(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))) (39)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 3#(5(4(3(1(0(1(x1))))))) (40)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 3#(3(5(4(3(1(0(1(x1)))))))) (41)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 3#(3(3(5(4(3(1(0(1(x1))))))))) (42)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 3#(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))) (43)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 3#(2(2(3(3(3(5(4(3(1(0(1(x1)))))))))))) (44)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 3#(1(0(1(x1)))) (45)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 2#(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) (46)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 2#(3(3(3(5(4(3(1(0(1(x1)))))))))) (47)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 2#(2(3(3(3(5(4(3(1(0(1(x1))))))))))) (48)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 1#(x1) (49)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 1#(0(1(x1))) (50)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 3#(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))) (51)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 3#(5(2(2(1(2(0(4(1(4(5(x1))))))))))) (52)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 2#(2(1(2(0(4(1(4(5(x1))))))))) (53)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 2#(1(2(0(4(1(4(5(x1)))))))) (54)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 2#(0(4(1(4(5(x1)))))) (55)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 1#(4(5(x1))) (56)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 1#(2(0(4(1(4(5(x1))))))) (57)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 3#(4(5(x1))) (58)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 3#(2(2(3(4(5(x1)))))) (59)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 2#(3(4(5(x1)))) (60)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 2#(3(2(2(3(4(5(x1))))))) (61)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 2#(2(3(4(5(x1))))) (62)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 2#(1(2(3(2(2(3(4(5(x1))))))))) (63)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 1#(2(3(2(2(3(4(5(x1)))))))) (64)
1#(2(1(0(0(x1))))) → 1#(x1) (65)
1#(2(1(0(0(x1))))) → 1#(1(0(1(x1)))) (66)
1#(2(1(0(0(x1))))) → 1#(0(1(x1))) (67)

1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[5(x1)] = x1 +
1
[4(x1)] = x1 +
1
[3(x1)] = x1 +
1
[2(x1)] = x1 +
1
[1(x1)] = x1 +
1
[0(x1)] = x1 +
1
[3#(x1)] = x1 +
0
[2#(x1)] = x1 +
1
[1#(x1)] = x1 +
1
together with the usable rules
1(2(1(0(0(x1))))) → 0(1(1(0(1(x1))))) (26)
1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 0(4(2(1(2(3(2(2(3(4(5(x1))))))))))) (27)
2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 2(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) (28)
2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 4(4(3(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))))) (29)
3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 0(3(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))))) (30)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 3#(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))))))) (31)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 3#(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))) (32)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 2#(5(5(0(x1)))) (33)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 2#(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))) (34)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 2#(5(2(4(2(5(0(2(5(5(0(x1))))))))))) (35)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 2#(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))))) (36)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 2#(5(0(2(5(5(0(x1))))))) (37)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 2#(4(2(5(0(2(5(5(0(x1))))))))) (38)
3#(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) → 1#(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))) (39)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 3#(5(4(3(1(0(1(x1))))))) (40)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 3#(3(5(4(3(1(0(1(x1)))))))) (41)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 3#(3(3(5(4(3(1(0(1(x1))))))))) (42)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 3#(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))) (43)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 3#(2(2(3(3(3(5(4(3(1(0(1(x1)))))))))))) (44)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 3#(1(0(1(x1)))) (45)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 2#(3(3(3(5(4(3(1(0(1(x1)))))))))) (47)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 2#(2(3(3(3(5(4(3(1(0(1(x1))))))))))) (48)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 1#(x1) (49)
2#(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) → 1#(0(1(x1))) (50)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 3#(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))) (51)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 3#(5(2(2(1(2(0(4(1(4(5(x1))))))))))) (52)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 2#(2(1(2(0(4(1(4(5(x1))))))))) (53)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 2#(1(2(0(4(1(4(5(x1)))))))) (54)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 2#(0(4(1(4(5(x1)))))) (55)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 1#(4(5(x1))) (56)
2#(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) → 1#(2(0(4(1(4(5(x1))))))) (57)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 3#(4(5(x1))) (58)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 3#(2(2(3(4(5(x1)))))) (59)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 2#(3(4(5(x1)))) (60)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 2#(3(2(2(3(4(5(x1))))))) (61)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 2#(2(3(4(5(x1))))) (62)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 2#(1(2(3(2(2(3(4(5(x1))))))))) (63)
1#(5(1(1(5(4(3(2(4(0(4(x1))))))))))) → 1#(2(3(2(2(3(4(5(x1)))))))) (64)
1#(2(1(0(0(x1))))) → 1#(x1) (65)
1#(2(1(0(0(x1))))) → 1#(1(0(1(x1)))) (66)
1#(2(1(0(0(x1))))) → 1#(0(1(x1))) (67)
and no rules could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.