I am making an attempt to make use of these Logical equivalence as axioms to show some PL statements,

On this case I adopted the examples within the documentation that did not use the build-in logic features${textual content{And}[,],textual content{Or}[,],textual content{and so on}}$, the reason being, in any other case it’ll consider some axioms as True.

Right here is my code: (Up to date)

```
PL = {ForAll[{p, q, r}, and[p, True] == p],
ForAll[{p, q, r}, or[p, False] == p],
ForAll[{p, q, r}, or[p, True] == True],
ForAll[{p, q, r}, and[p, False] == False],
ForAll[{p, q, r}, or[p, p] == p],
ForAll[{p, q, r}, and[p, p] == p],
ForAll[{p, q, r}, not[not[p]] == p],
ForAll[{p, q, r}, or[p, q] == or[q, p]],
ForAll[{p, q, r}, and[p, q] == and[q, p]],
ForAll[{p, q, r}, or[or[p, q], r] == or[p, or[q, r]]],
ForAll[{p, q, r}, and[and[p, q], r] == and[p, and[q, r]]],
ForAll[{p, q, r}, or[p, and[q, r]] == and[or[p, q], or[p, r]]],
ForAll[{p, q, r}, and[p, or[q, r]] == or[and[p, q], and[p, r]]],
ForAll[{p, q, r}, not[and[p, q]] == or[not[p], not[q]]],
ForAll[{p, q, r}, not[or[p, q]] == and[not[p], not[q]]],
ForAll[{p, q, r}, or[p, and[p, q]] == p],
ForAll[{p, q, r}, and[p, or[p, q]] == p],
ForAll[{p, q, r}, or[p, not[p]] == True],
ForAll[{p, q, r}, and[p, not[p]] == False]}
proof = FindEquationalProof[
ForAll[{p, q, r},
and[or[and[and[r, not[q]], not[p]],
and[q, not[and[q, not[or[p, r]]]]]],
not[and[p, not[and[q, not[r]]]]]] ==
or[and[and[p, q], not[and[r, q]]], and[r, not[p]]]], PL]
proof["ProofGraph"]
proof["ProofNotebook"]
```

I simply fastened the typo within the axiom, but making an attempt to let it show that assertion:

$$((r ∧ ¬q ∧ ¬p) ∨ (q ∧ ¬(q ∧ ¬(p ∨ r)))) ∧ ¬(p ∧ ¬(q ∧ ¬r))$$

$$equiv(p ∧ q ∧ ¬(r ∧ q)) ∨ (r ∧ ¬p)$$

```
and[or[and[and[r, not[q]], not[p]],
and[q, not[and[q, not[or[p, r]]]]]],
not[and[p, not[and[q, not[r]]]]]] ==
or[and[and[p, q], not[and[r, q]]], and[r, not[p]]]], PL]
```

However appears not work, i attempted shorter ones, which works nice, is it as a result of this assertion too lengthy or one thing I missed $?$

Nonetheless it may be proved with booleanLogic axioms given in documentation, with 402 steps:

```
booleanLogic = {ForAll[{a, b}, and[a, b] == and[b, a]],
ForAll[{a, b}, or[a, b] == or[b, a]],
ForAll[{a, b}, and[a, or[b, not[b]]] == a],
ForAll[{a, b}, or[a, and[b, not[b]]] == a],
ForAll[{a, b, c}, and[a, or[b, c]] == or[and[a, b], and[a, c]]],
ForAll[{a, b, c}, or[a, and[b, c]] == and[or[a, b], or[a, c]]]}
proof = FindEquationalProof[
ForAll[{p, q, r},
and[or[and[and[r, not[q]], not[p]],
and[q, not[and[q, not[or[p, r]]]]]],
not[and[p, not[and[q, not[r]]]]]] ==
or[and[and[p, q], not[and[r, q]]], and[r, not[p]]]], booleanLogic]
proof["ProofGraph"]
proof["ProofNotebook"]
```

And right here is the proof graph:

Any assist can be appreciated.