Does $P\iff S$ follow from $P\iff Q,P\to\lnot R,P\to S,R\to S,S\to(P\lor R)$, and $\lnot P \rightarrow S$ is false?

Question

I am working on proving a theorem and just want to make sure my conclusion is true. Suppose we are given the statements

\begin{equation}\tag{1} P \iff Q \end{equation}

\begin{equation}\tag{2} P \rightarrow \lnot R \end{equation}

\begin{equation}\tag{3} P \rightarrow S \end{equation}

\begin{equation}\tag{4} R \rightarrow S \end{equation}

\begin{equation}\tag{5} S \rightarrow ( P \lor R) \end{equation}

Now suppose that it is also known that the contrapositive of $\lnot P \rightarrow S$ is false, meaning that $\lnot P \rightarrow S$ is false when $\lnot P$ is true. My goal is to show that

\begin{equation} P \iff S \end{equation}

To do this, it is clear that when $S$ is true, it follows that $(P \lor R)$ is true. Since $(P \lor R)$ is the same as $\lnot P \rightarrow R$, the conjunction of equation $4$ and $\lnot P \rightarrow R$ produces $\lnot P \rightarrow S$ through hypothetical syllogism, which must be true when $S$ is true. Since this statement was derived from statements that are true, the only way $\lnot P \rightarrow S$ is true is when $\lnot P$ is false. By double negation, it follows that when $S$ is true, $\lnot \lnot P$ is true, meaning $P$ is true. This, along with equation 3 shows

\begin{equation} P \iff S \end{equation}

I just want to make sure there is nothing circular about what I have done.

Answer 1

To claim that the conclusion follows the premises, you have to show that $P \leftrightarrow S$ is true whenever all the premises $(1){-}(5)$ are simultaneously true.

Now, consider the situation where $S$ and $R$ are true, and $P$ and $Q$ are false. Then the conclusion $P \leftrightarrow S$ is false, but all the premises are true, indeed:

1. $P \leftrightarrow Q$ is true because both $P$ and $Q$ are false;
2. $P \to \lnot R$ is true because $P$ is false;
3. $P \to S$ is true because $P$ is false;
4. $R \to S$ is true because $S$ is true;
5. $S \to (P \lor R)$ is true because $P \lor R$ is true.

Therefore, the conclusion $P \leftrightarrow S$ does not follow from the premises $(1) {-} (5)$.

Answer 2

Does $P\iff S$ follow from $P\iff Q,P\to\lnot R,P\to S,R\to S,S\to(P\lor R)$, and $\lnot P \rightarrow S$ is false?

Interpreting as:

$(P\iff Q) \land (P\implies \lnot R) \land (P\implies S) \land (R\implies S)\land (S\implies (P\lor R))\land \lnot (\lnot P \rightarrow S)\\\implies (P\iff S)$

This is a tautology. Here is the truth table:

Simplfied:

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