Certification Problem

Input (COPS 116)

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 → b7 (24)
b7 → b1 (25)
b7 → c1 (26)

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}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2022)

1 Redundant Rules Transformation

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

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 → c4 (14)
b4 → b5 (15)
c4 → c5 (16)
a5 → c5 (18)
b5 → b6 (19)
c5 → c6 (20)
a6 → c6 (22)
b6 → b7 (23)
c6 → b7 (24)
b7 → b1 (25)
b7 → c1 (26)

All redundant rules that were added or removed can be simulated in 4 steps .

1.1 Strongly closed

Confluence is proven since the TRS is strongly closed. The joins can be performed using 7 step(s).