سوال ۹۱ کنکور ارشد علوم کامپیوتر ۱۴۰۳

Test each argument for validity: an argument is valid iff every interpretation that makes all premises true also makes the conclusion true. Finding a single counter-model (premises true, conclusion false) proves invalidity.

Argument (a)

With predicates $R(x)$ = "$x$ is rational", $\text{Re}(x)$ = "$x$ is real", $C(x)$ = "$x$ is complex":

$\forall x\,(R(x)\to \text{Re}(x)),\qquad \exists x\,(C(x)\wedge\neg R(x))\ \vdash\ \exists x\,(C(x)\wedge\neg \text{Re}(x)).$

This has the syllogistic shape All $R$ are $\text{Re}$; some $C$ are not $R$; therefore some $C$ are not $\text{Re}$, which is a classic invalid form (the middle term $R$ does not connect $C$ to $\text{Re}$ the right way). A counter-model: let every object be real, let the rationals be a subset of the reals, and let the "complex" objects be exactly the irrational reals. Then both premises hold (all rationals are real; the irrational "complex" objects are not rational), yet no complex object is non-real, so the conclusion is false. Hence (a) is invalid.

Argument (b)

With $p$ = "I passed", $m$ = "good midterm", $f$ = "good final":

$p\to m,\qquad p\to f,\qquad m\wedge f\ \vdash\ p.$

This is the fallacy of affirming the consequent. Take the assignment

$p=\text{F},\quad m=\text{T},\quad f=\text{T}.$

Then $p\to m=\text{T}$, $p\to f=\text{T}$, and $m\wedge f=\text{T}$ — all three premises are true — but the conclusion $p=\text{F}$. A counterexample exists, so (b) is invalid.

Conclusion

Both arguments are invalid, so the answer is neither (option 4).

Pitfall: Premises and conclusion can all be true facts about the world and the argument still be invalid — validity is about the form (no counter-model), not about whether the sentences happen to be true. Conversely, true-looking premises plus a believable conclusion (as in (b)) is exactly how affirming-the-consequent slips past you.