Can we have $(a \wedge b) \to c$ but not $a \to c$ or $b \to c$?

Question

I could like to know can we have:

• $a \not \to c$
• $b \not \to c$
• however $(a \wedge b) \to c$

what is a logic table fitting that criterion? In general:

• $a_k \not \to c$ for any $k = 1, 2, \dots, n$
• yet we observe: $(a_1 \wedge \dots \wedge a_n) \to c$

Answer 1

Yes, consider "I have my username" = a and "I have my password" = b. Then let c = "I can logon to my account."

More generally this can occur quite easily as well, trivially you could have a vector of $n$ entries and let $a_n$ = "The nth letter is [letter here]" and c = "the vector is equal to [vector here]."

Answer 2

No. I'll try to work this out to show to show via a truth-table (with the help of wolfram alpha) that the conjunction of all three following propositions cannot hold true, simultaneously:

$$(A\land B) \rightarrow C \equiv \lnot(A\land B) \lor C\tag 1$$

$$\lnot (B\rightarrow C) \equiv \lnot (\lnot B \lor C)\tag 2$$ $$\lnot (A\rightarrow C) \equiv \lnot (\lnot B \lor C)\tag 3$$

So $(A\land B)\rightarrow C$ cannot hold true given ($\lnot (A\rightarrow C)$ and $\lnot (B\rightarrow C).$)

Answer 3

You’re asking if $(a\land b)\to c$ and $\lnot(a\to c)\land\lnot(b\to c)$ can both be true at the same time. Let’s rewrite the first formula: $$\begin{align}(a\land b)\to c &\equiv \lnot(a\land b)\lor c \\ &\equiv \lnot a\lor\lnot b\lor c \\ &\equiv (\lnot a\lor c)\lor(\lnot b\lor c) \\ &\equiv (a\to c)\lor(b\to c).\end{align}$$ That last formula is the negation of $\lnot(a\to c)\land\lnot(b\to c)$, so the latter clearly can’t be true when $(a\land b)\to c$ is. Indeed, we can see directly from the last equivalence that if the first implication is true, then at least one of the individual implications must be as well.

Answer 4

It depends... From a propositional logic point of view, if 'a', 'b' and 'c' are simple propositions, it is not possible; however, if 'a', 'b', and 'c' are not necessarily simple, then it could be the case that 'a' = p → q, 'b' = p, and 'c' = q. In this case, it´s correct to say that ¬(a→c)∧¬(b→c) and (a∧b)→c.

Moreover, from a predicate logic point of view, it is obviously true (in some cases). For example: ¬(∀x (Px → Qx) → Qa) and ¬(Pa→Qa) but (∀x (Px → Qx)∧ Pa)→Qa.