If I want to show (A and B) imply C with a contrapositive proof, can I assume (B and ~C) and show ~A? If I want to show (A and B) imply C with a contrapositive proof, can I assume (B and ~C) and show ~A?
Is showing (~C and B) implies ~A the same as showing ~C implies ~(A and B)?
Thanks in advance.
If you assume $\lnot C$, then $\lnot C \to [\lnot(A \land B)] \equiv (\lnot A \lor \lnot B)$, by the contrapositve.
If you also know that $B$ holds, then you can conclude that this in turn implies that $\lnot A$ holds, by disjunction elimination, hence you can conclude &quot;If $(\lnot C \land B) \to \lnot A$&quot;.  But that only proves
$(1) \; [(A\land B) \to C]\quad \text{premise}$
$(2) B \;\quad \text{premise}$
$(3)\qquad \lnot C \quad \text{Assumption}$
$(4)\qquad\quad \lnot (A\land B) \quad \text{Modus Tollens, 1, 3}$
$(5)\qquad\quad \lnot A \lor \lnot B\quad \text{DeMorgan's, 4}$
$(6) \qquad\quad \lnot\lnot B\quad \text{Double Negation, 2}$
$(7) \qquad\quad \lnot A \quad \text{Disjunctive Syllogism, 5, 6}$
$(8) \qquad\quad\lnot A \lor \lnot B \quad \lor\text{-Introduction}$
$(9) \lnot C \to (\lnot A\lor \lnot B) \quad \text{conditional introduction}$
$\therefore (10) \lnot C \to \lnot (A \land B)\quad \text{DeMorgan's}$
And $$ \lnot C \to \lnot (A \land B) \equiv (A\land B)\to C$$ as its contrapositive, hence the proof is complete.


Perhaps more straight forward for your title question:  Yes.
This requires you understand the equivalence: $$(p \land q) \to r \equiv p \to (q\to r).$$
In your example, that means $(A\land B) \to C \equiv A \to (B\to C).\tag{1}$$
Now (1) is equivalent to its contrapostive:
$$\lnot(B \to C) \to \lnot A \equiv \lnot (\lnot B \lor C)\to \lnot A \equiv (B\land \lnot C)\to \lnot A\tag{2}$$
So with (2), assuming $B \land \lnot C$, then indeed, by Modus Ponens, $\lnot A$ is implied.
By contrapositive equivalence, you hav
If you assume $\lnot C$, then $\lnot C \to [\lnot(A \land B)] \equiv (\lnot A \lor \lnot B)$, by the contrapositve.
If you also know that $B$ holds, then you can conclude that this in turn implies that $\lnot A$ holds, by disjunction elimination, hence you can conclude &quot;If $(\lnot C \land B) \to \lnot A$&quot;.  But that only proves
$(1) \; [(A\land B) \to C]\quad \text{premise}$
$(2) B \;\quad \text{premise}$
$(3)\qquad \lnot C \quad \text{Assumption}$
$(4)\qquad\quad \lnot (A\land B) \quad \text{Modus Tollens, 1, 3}$
$(5)\qquad\quad \lnot A \lor \lnot B\quad \text{DeMorgan's, 4}$
$(6) \qquad\quad \lnot\lnot B\quad \text{Double Negation, 2}$
$(7) \qquad\quad \lnot A \quad \text{Disjunctive Syllogism, 5, 6}$
$(8) \qquad\quad\lnot A \lor \lnot B \quad \lor\text{-Introduction}$
$(9) \lnot C \to (\lnot A\lor \lnot B) \quad \text{conditional introduction}$
$\therefore (10) \lnot C \to \lnot (A \land B)\quad \text{DeMorgan's}$
And $$ \lnot C \to \lnot (A \land B) \equiv (A\land B)\to C$$ as its contrapositive, hence the proof is complete.


Perhaps more straight forward for your title question:  Yes.
This requires you understand the equivalence: $$(p \land q) \to r \equiv p \to (q\to r).$$
In your example, that means $(A\land B) \to C \equiv A \to (B\to C).\tag{1}$$
Now (1) is equivalent to its contrapostive:
$$\lnot(B \to C) \to \lnot A \equiv \lnot (\lnot B \lor C)\to \lnot A \equiv (B\land \lnot C)\to \lnot A\tag{2}$$
So with (2), assuming $B \land \lnot C$, then indeed, by Modus Ponens, $\lnot A$ is implied.
By contrapositive equivalence, you hav