First, I have defined a well-formed formula as such:
1) Each atom is a WFF.
2) If φ is a WFF, so is ¬φ
3) If φ and ψ are WFFs, if ∗ is a binary connected (i.e., ∨,∧,→), then (φ∗ψ) is a WFF.
What I have done so far:
Let ϕ be a well formed formula and we will show that every two atoms in ϕ have a connective between them. We can say that for every i < j , if ai and aj are propositional variables in ϕ, then there is some i < k < j such than the symbol ak is a connective.
Now, for the base case, if ϕ is an atom itself, then there simply cannot exist two different atoms ai and aj such that ai and aj are both atoms. Thus, the base case vacuously holds true.
For the induction step,we assume α = a1 ...an and β = b1... bm satisfy the property that there exits a connective between each pair of atoms.
We have 2 cases to prove.
1) ϕ = ( ¬α )
2) ϕ = (α∗β) (∗∈(∨,∧,→)
This induction step is where I am stuck. Perhaps there is a simpler approach that I am overlooking but any guidance would be very much appreciated.
Thanks in advance!
1) Let $p$ and $q$ be two atoms in $\lnot \alpha$. Then $p$ and $q$ are two atoms in $\alpha$. By the induction hypothesis, there is a connective $k$ in $\alpha$ between $p$ and $q$. This $k$ is a connective in $\lnot\alpha$ between $p$ and $q$.
2) Let $p$ and $q$ be two atoms in $\alpha\ast \beta$. If $p$ and $q$ are in $\alpha$, by the induction hypothesis there is a connective between $p$ and $q$. The same holds if $p$ and $q$ are in $\beta$. In either case there is a connective in $\alpha\ast\beta$ between $p$ and $q$.
Finally suppose that $p$ is in one of $\alpha$ or $\beta$, say in $\alpha$, and $q$ is in $\beta$. Then $\ast$ is a connective in $\alpha\ast\beta$ between $p$ and $q$.