Certification Problem

Input (COPS 115)

We consider the TRS containing the following rules:

a1 → b1 (1)
a1 → c1 (2)
b1 → b2 (3)
c1 → c2 (4)
a2 → b2 (5)
a2 → c2 (6)
b2 → b3 (7)
c2 → c3 (8)
a3 → b3 (9)
a3 → c3 (10)
b3 → b4 (11)
c3 → c4 (12)
a4 → b4 (13)
a4 → c4 (14)
b4 → b5 (15)
c4 → c5 (16)
a5 → b5 (17)
a5 → c5 (18)
b5 → b6 (19)
c5 → c6 (20)
a6 → b6 (21)
a6 → c6 (22)
b6 → b7 (23)
c6 → c7 (24)
a7 → b7 (25)
a7 → c7 (26)
b7 → b8 (27)
c7 → c8 (28)
a8 → b8 (29)
a8 → c8 (30)
b8 → b9 (31)
c8 → c9 (32)
a9 → b9 (33)
a9 → c9 (34)
b9 → b10 (35)
c9 → c10 (36)
a10 → b11 (37)
b10 → b11 (38)
c10 → b11 (39)

The underlying signature is as follows:

{a1/0, b1/0, c1/0, b2/0, c2/0, a2/0, b3/0, c3/0, a3/0, b4/0, c4/0, a4/0, b5/0, c5/0, a5/0, b6/0, c6/0, a6/0, b7/0, c7/0, a7/0, b8/0, c8/0, a8/0, b9/0, c9/0, a9/0, b10/0, c10/0, a10/0, b11/0}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2020)

1 Redundant Rules Transformation

To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:

c10 → b11 (39)
b10 → b11 (38)
a10 → b11 (37)
c9 → c10 (36)
b9 → b10 (35)
a9 → c9 (34)
a9 → b9 (33)
c8 → c9 (32)
b8 → b9 (31)
a8 → c8 (30)
a8 → b8 (29)
c7 → c8 (28)
b7 → b8 (27)
a7 → c7 (26)
a7 → b7 (25)
c6 → c7 (24)
b6 → b7 (23)
a6 → c6 (22)
a6 → b6 (21)
c5 → c6 (20)
b5 → b6 (19)
a5 → c5 (18)
a5 → b5 (17)
c4 → c5 (16)
b4 → b5 (15)
a4 → c4 (14)
a4 → b4 (13)
c3 → c4 (12)
b3 → b4 (11)
a3 → c3 (10)
a3 → b3 (9)
c2 → c3 (8)
b2 → b3 (7)
a2 → c2 (6)
a2 → b2 (5)
c1 → c2 (4)
b1 → b2 (3)
a1 → c1 (2)
a1 → b1 (1)
b1 → b11 (40)
c1 → b11 (41)
b2 → b11 (42)
c2 → b11 (43)
b3 → b11 (44)
c3 → b11 (45)
b4 → b11 (46)
c4 → b11 (47)
b5 → b11 (48)
c5 → b11 (49)
b6 → b11 (50)
c6 → b11 (51)
b7 → b11 (52)
c7 → b11 (53)
b8 → b11 (54)
c8 → b11 (55)
b9 → b11 (56)
c9 → b11 (57)

All redundant rules that were added or removed can be simulated in 10 steps .

1.1 Parallel Closed

Confluence is proven since the TRS is (almost) parallel closed. The joins can be performed using 1 parallel step(s).