# All babies are querulous. But David is not a baby, so he is not querulous [closed]

Does the following argument involve a fallacy?

All babies are querulous. But David is not a baby, so he is not querulous.

(a) undistributed middle

(b) denying the antecedent

(c) affirming the consequent

(d) no fallacy; deductively valid

## closed as off-topic by Chris Sunami, John Am, Joseph Weissman♦Sep 7 '16 at 20:35

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• From previous readings, denying the antecedent would be "if this, then that. But not this, so that" In this : P1- All babies are querulous. P2- David is not a baby. C- So, David is not querulous. Therefore B, is seemingly the fallacy. – Nicola Sep 7 '16 at 4:34
• You're right, this is indeed denying the antecedent. – commando Sep 7 '16 at 4:59
• If the question were altered to : " All babies are querulous. David is not querulous, so he is not a baby" Is it the variant: affirming the consequent ? @commando – Nicola Sep 7 '16 at 5:05
• We can look forward to much more site traffic in the coming weeks/months if you're going to post every homework question this term :) – Keelan Sep 7 '16 at 5:53
• Would this be such a bad thing @Keelan ? More input and feedback to questions each and every individual has, broadening ones knowledge? Sorry if you are frustrated! I beg my pardon for asking for help! – Nicola Sep 7 '16 at 6:24

Yes, the given argument involves a formal fallacy, denying the antecedent, which goes like this:

1. p → q
2. ~p
3. Therefore, ~q

The conclusion doesn't logically follow from the premises.

The given argument is clearly an instance of that:

1. ∀x(Bx → Qx) [all babies are querulous]
2. ~Bd [David is not a baby]
3. Therefore, ~Qd [David is not querulus]