I am having a problem understanding a proof from the field of mathematical logic. Seems like my brain cannot digest concepts from logic very well.
I will quickly define some terminology and then present my issue.
We say that a theory T is contradictory if there is $ \vdash_T \perp$ that is, there exist a proof of $\perp.$ We already proved that propositional logic $\mathbb{I}$ is not contradictory. Let $\mathbb{P}$ denote first order logic without the rules for equality.
I want to show that $\mathbb{P}$ is also not contradictory. The idea we used at the classes is to construct a function $s$ mapping formulas from $\mathbb{P}$ to formulas in $\mathbb{I}$ such that
- $s(\perp) = \perp$
- If $ \vdash_{\mathbb{P}} \phi$ then $\vdash_{\mathbb{I}} s(\phi)$ that is, if there's a proof of $\phi$ using the rules of $\mathbb{P}$ then there is a proof of $s(\phi)$ in $\mathbb{I}$.
The existence of such a $s$ would then imply that $\mathbb{P}$ is not contradictory.
In our classes we defined $s$ in a certain manner and showed that $1.$ and $2.$ holds by ways of induction on the structure of a possible formula in $\mathbb{P}.$
I am wondering why the function defined as $s(\perp) = \perp$ and $s(x) = \top$ if $ x \not = \perp$ wouldn't be good as well?
I am aware this must be quite a dumb proposal since in this case, conditions 1. and 2. are satisfied blindly (are they?) and there is no difference in taking $\mathbb{P}$ or simply taking the whole first order theory with rules for equality included.
So either there is something else that $s$ must satisfy in order to be able to conclude that such an $s$ guarantees that $\mathbb{P}$ is not contradictory, "my" $s$ does not satisfy the assumptions 1. 2. or some other stupid confusion I am having.
Can someone enlighten me?