5. Proposition: WFFs must be unambiguous.

Proof: Every WFF is either an atomic formula, has a unary connective, and/or a binary connective. Moreover, without the proper parenthesis, it is unclear in which order the operations in a WFF should be conducted. Formulas like ϕ∧ψ∨θ can produce different results depending on who is reading the formula. (ϕ∧(ψ∨θ)) is read as ϕ is always true and either ψ is true and θ or θ is true and ψ is false. ((ϕ∧ψ)∨θ) is read as either ϕ and ψ are always true and θ is false or θ is true. These are clearly two different results. Unique readability means that this confusion will not arise as the formula should be clear enough that anyone reading it would arrive at the same conclusion.

Assume that this holds for all WFFs, a complete WFF has unique readability. Without a complete WFF, it is unclear what operation is being performed. ∀x(x = y) is a complete WFF with ∀x as an initial segment. The initial segment ∀x, does not yield a true or false answer like a full WFF would. Therefore, complete WFFs satisfy the conditions for unique readability.

6. Proposition: There exist irrational numbers A, B such that A^{B} is rational.

Proof: We know that $\sqrt{2}$ is irrational. So, if A=$\sqrt{2}$ and B =$\sqrt{2}$ satisfy the theorem, then we are done. If they do not, then $\sqrt{2}$^{$\sqrt{2}$} is irrational, so let A now be this number. Then, letting B=$\sqrt{2}$, it is easy to verify that A^{B} = 2 which is rational and hence would satisfy the theorem.

($\sqrt{2}$^{$\sqrt{2}$})^{$\sqrt{2}$} = $\sqrt{2}$^{$2$} = 2

You cannot use this proof to find specific values of a and b, it can only be used to verify that the two values satisfy the theorem. There is nothing in the proof that would return any values for A and B.

Comments - Andrew Furash

You don't actually prove anything in 5. What you need is a characterization of 'unique readability.' That is, one thing that every unique formula has and no non-unique formula has or something every non-unique formula has and no unique formula has. Then, you can use the lemma to show that WFFs satisfy the appropriate characteristic.

6 is great.