Certificate for #12591 ⟨a, b | abba=ba, bbaa=b

Completion settings:

[1] abba=ba

Axiom: abba=ba.

Referenced by [3], [4], [5], [6], [7], [9].

[2] bbaa=b

Axiom: bbaa=b.

Referenced by [4], [8], [11], [13].

[3] abbba=bba

Overlap of [1] abba=ba with [1] abba=ba:

abb a abba

Critical pair: abbba=babba.

Reduce RHS:

[1]b(abba)
bba

Referenced by [6].

[4] baa=ab

Overlap of [1] abba=ba with [2] bbaa=b:

a bba bbaa

Critical pair: ab=baa.

Flip LHS and RHS.

Referenced by [5], [6], [9], [10], [12], [13], [14].

[5] abab=ab

Overlap of [1] abba=ba with [4] baa=ab:

ab ba baa

Critical pair: abab=baa.

Reduce RHS:

[4](baa)
ab

Referenced by [7], [8], [9], [10], [12].

[6] baba=bba

Overlap of [4] baa=ab with [1] abba=ba:

ba a abba

Critical pair: baba=abbba.

Reduce RHS:

[3](abbba)
bba

Referenced by [7].

[7] bbab=bab

Overlap of [1] abba=ba with [5] abab=ab:

abb a abab

Critical pair: abbab=babab.

Reduce LHS:

[1](abba)b
bab

Reduce RHS:

[6](baba)b
bbab

Flip LHS and RHS.

Referenced by [8], [11].

[8] bab=bb

Overlap of [2] bbaa=b with [5] abab=ab:

bba a abab

Critical pair: bbaab=bbab.

Reduce LHS:

[2](bbaa)b
bb

Reduce RHS:

[7](bbab)
bab

Flip LHS and RHS.

Referenced by [9], [11], [12].

[9] abb=bb

Overlap of [4] baa=ab with [5] abab=ab:

ba a abab

Critical pair: baab=abbab.

Reduce LHS:

[4](baa)b
abb

Reduce RHS:

[1](abba)b
[8](bab)
bb

Referenced by [10].

[10] aab=bb

Overlap of [5] abab=ab with [4] baa=ab:

aba b baa

Critical pair: abaab=abaa.

Reduce LHS:

[4]a(baa)b
[9]a(abb)
[9](abb)
bb

Reduce RHS:

[4]a(baa)
aab

Flip LHS and RHS.

Referenced by [11], [12], [13].

[11] bbb=bb

Overlap of [2] bbaa=b with [10] aab=bb:

bba a aab

Critical pair: bbabb=bab.

Reduce LHS:

[7](bbab)b
[8](bab)b
bbb

Reduce RHS:

[8](bab)
bb

Referenced by [12].

[12] bb=ab

Overlap of [4] baa=ab with [10] aab=bb:

ba a aab

Critical pair: babb=abab.

Reduce LHS:

[8](bab)b
[11](bbb)
bb

Reduce RHS:

[5](abab)
ab

Referenced by [13], [15].

[13] ab=b

Overlap of [2] bbaa=b with [12] bb=ab:

bbaa bb

Critical pair: abaa=b.

Reduce LHS:

[4]a(baa)
[10](aab)
[12](bb)
ab

Defines rule #1.

Referenced by [14], [15].

[14] baa=b

Simplify [4] baa=ab.

Reduce RHS:

[13](ab)
b

Defines rule #3.

[15] bb=b

Simplify [12] bb=ab.

Reduce RHS:

[13](ab)
b

Defines rule #2.