| Back: | ⟨a, b | aab=ba, bab=ab⟩ |
|---|
Completion settings:
Axiom: aab=ba.
Referenced by [3], [4], [5], [6], [7], [8].
Axiom: bab=ab.
Referenced by [3], [4], [6], [7].
Overlap of [1] aab=ba with [2] bab=ab:
Critical pair: aaab=baab.
Reduce LHS:
| [1] | a(aab) |
| ⇒ aba |
Reduce RHS:
| [1] | b(aab) |
| ⇒ bba |
Flip LHS and RHS.
Overlap of [2] bab=ab with [2] bab=ab:
Critical pair: baab=abab.
Reduce LHS:
| [1] | b(aab) |
| [3] | ⇒ (bba) |
| ⇒ aba |
Reduce RHS:
| [2] | a(bab) |
| [1] | ⇒ (aab) |
| ⇒ ba |
Overlap of [1] aab=ba with [4] aba=ba:
Critical pair: aba=baa.
Reduce LHS:
| [4] | (aba) |
| ⇒ ba |
Flip LHS and RHS.
Referenced by [6].
Overlap of [4] aba=ba with [1] aab=ba:
Critical pair: abba=baab.
Reduce LHS:
| [3] | a(bba) |
| [1] | ⇒ (aab)a |
| [5] | ⇒ (baa) |
| ⇒ ba |
Reduce RHS:
| [5] | (baa)b |
| [2] | ⇒ (bab) |
| ⇒ ab |
Defines rule #1.
Overlap of [4] aba=ba with [2] bab=ab:
Critical pair: aab=bab.
Reduce LHS:
| [1] | (aab) |
| [6] | ⇒ (ba) |
| ⇒ ab |
Reduce RHS:
| [6] | (ba)b |
| ⇒ abb |
Flip LHS and RHS.
Defines rule #3.
Simplify [1] aab=ba.
Reduce RHS:
| [6] | (ba) |
| ⇒ ab |
Defines rule #2.