| Back: | ⟨a, b | abba=ba, bbaa=b⟩ |
|---|
Completion settings:
Axiom: abba=ba.
Referenced by [3], [4], [5], [6], [7], [9].
Axiom: bbaa=b.
Referenced by [4], [8], [11], [13].
Overlap of [1] abba=ba with [1] abba=ba:
Critical pair: abbba=babba.
Reduce RHS:
| [1] | b(abba) |
| ⇒ bba |
Referenced by [6].
Overlap of [1] abba=ba with [2] bbaa=b:
Critical pair: ab=baa.
Flip LHS and RHS.
Referenced by [5], [6], [9], [10], [12], [13], [14].
Overlap of [1] abba=ba with [4] baa=ab:
Critical pair: abab=baa.
Reduce RHS:
| [4] | (baa) |
| ⇒ ab |
Referenced by [7], [8], [9], [10], [12].
Overlap of [4] baa=ab with [1] abba=ba:
Critical pair: baba=abbba.
Reduce RHS:
| [3] | (abbba) |
| ⇒ bba |
Referenced by [7].
Overlap of [1] abba=ba with [5] abab=ab:
Critical pair: abbab=babab.
Reduce LHS:
| [1] | (abba)b |
| ⇒ bab |
Reduce RHS:
| [6] | (baba)b |
| ⇒ bbab |
Flip LHS and RHS.
Overlap of [2] bbaa=b with [5] abab=ab:
Critical pair: bbaab=bbab.
Reduce LHS:
| [2] | (bbaa)b |
| ⇒ bb |
Reduce RHS:
| [7] | (bbab) |
| ⇒ bab |
Flip LHS and RHS.
Referenced by [9], [11], [12].
Overlap of [4] baa=ab with [5] abab=ab:
Critical pair: baab=abbab.
Reduce LHS:
| [4] | (baa)b |
| ⇒ abb |
Reduce RHS:
| [1] | (abba)b |
| [8] | ⇒ (bab) |
| ⇒ bb |
Referenced by [10].
Overlap of [5] abab=ab with [4] baa=ab:
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].
Overlap of [2] bbaa=b with [10] aab=bb:
Critical pair: bbabb=bab.
Reduce LHS:
| [7] | (bbab)b |
| [8] | ⇒ (bab)b |
| ⇒ bbb |
Reduce RHS:
| [8] | (bab) |
| ⇒ bb |
Referenced by [12].
Overlap of [4] baa=ab with [10] aab=bb:
Critical pair: babb=abab.
Reduce LHS:
| [8] | (bab)b |
| [11] | ⇒ (bbb) |
| ⇒ bb |
Reduce RHS:
| [5] | (abab) |
| ⇒ ab |
Overlap of [2] bbaa=b with [12] bb=ab:
Critical pair: abaa=b.
Reduce LHS:
| [4] | a(baa) |
| [10] | ⇒ (aab) |
| [12] | ⇒ (bb) |
| ⇒ ab |
Defines rule #1.
Simplify [4] baa=ab.
Reduce RHS:
| [13] | (ab) |
| ⇒ b |
Defines rule #3.
Simplify [12] bb=ab.
Reduce RHS:
| [13] | (ab) |
| ⇒ b |
Defines rule #2.