Given the propositions A, B and C, we assume that \begin{equation} (A\wedge B)\rightarrow C \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \end{equation}
I want to demonstrate the other way around implication
\begin{equation}
C \rightarrow (A\wedge B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\end{equation}
Obviously statement (1) doesn't necesssarely imply statement (2) anyway I started my strange reasoning
Attempted proof
by contraposition of the statement (1), we get
$$\neg C\rightarrow \neg A \vee \neg B \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$
and the proposition (2) implies the following statement
$$\neg C\vee(A\wedge B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$$ Now, here comes my doubt!
Combining statements (3) and (4) do I get the following statement?
$$(\neg A \vee \neg B)\vee(A\wedge B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5)$$
the intuition suggests that writing $\neg C$ or $\neg A \vee \neg B$ doesn't change anything in statement (4) since the implication (3) holds by assumption.
If yes, since the statement (5) is always true therefore the statement 2 would be true as well.
Am I right? Am I wrong?
Thank you!
Long comment
(1) holds by assumption and it is equivalent to (3) [and (4)].
We want to prove that also (2) holds.
What does it mean "combining statements (3) and (4)" ?
Conjunction ? If so, the "combination" implies (5): $(¬A∨¬B) ∨ (A∧B)$, and this formula is a tautology, i.e. always true.
Now we have the problem: "the statement (5) is always true therefore the statement (2) would be true as well." Why ?
We are trying to prove (2) and thus we are not entitled to assert that it is true.
The fact that (2) implies a True proposition does not mean that (2) is true: a tautology is implied by every proposition.
For example:
is a valid inference for a formula $\varphi$ whatever.
The short answer is that $(A∧B) → C \nvDash C → (A∧B)$.
We can check it with truth-table: for $C$ True and $A$ False, we have that (1) is True while (2) is False.