Certificate for #3422 ⟨a, b, c | ba=ab, aca=a⟩

Completion settings:

[1] ba=ab

Axiom: ba=ab.

Referenced by [4], [5], [8], [9].

[2] aca=a

Axiom: aca=a.

Defines rule #5.

Referenced by [4], [6].

[3] bca=d

Axiom: bca=d.

Defines rule #8.

Referenced by [4], [6], [7].

[4] ab=ad

Overlap of [1] ba=ab with [2] aca=a:

b a aca

Critical pair: ba=abca.

Reduce LHS:

[1](ba)
⇒ ab

Reduce RHS:

[3]a(bca)
⇒ ad

Defines rule #1.

Referenced by [5], [7], [8], [9].

[5] ada=aad

Overlap of [4] ab=ad with [1] ba=ab:

a b ba

Critical pair: aab=ada.

Reduce LHS:

[4]a(ab)
⇒ aad

Flip LHS and RHS.

Defines rule #4.

[6] dca=d

Overlap of [3] bca=d with [2] aca=a:

bc a aca

Critical pair: bca=dca.

Reduce LHS:

[3](bca)
⇒ d

Flip LHS and RHS.

Defines rule #7.

[7] db=dd

Overlap of [3] bca=d with [4] ab=ad:

bc a ab

Critical pair: bcad=db.

Reduce LHS:

[3](bca)d
⇒ dd

Flip LHS and RHS.

Defines rule #2.

Referenced by [8].

[8] dda=dad

Overlap of [7] db=dd with [1] ba=ab:

d b ba

Critical pair: dab=dda.

Reduce LHS:

[4]d(ab)
⇒ dad

Flip LHS and RHS.

Defines rule #6.

[9] ba=ad

Simplify [1] ba=ab.

Reduce RHS:

[4](ab)
⇒ ad

Defines rule #3.