The Logic Problem, taken from WFF’N PROOF, The
Game of Logic, has these two assumptions:
- “Logic is difficult or not many students like logic.”
- “If mathematics is easy, then logic is not difficult.”
By translating these assumptions into statements involving
propositional variables and logical connectives, determine
whether each of the following are valid conclusions
of these assumptions:
- a) That mathematics is not easy, if many students like logic.
- b) That not many students like logic, if mathematics is not easy.
- c) That mathematics is not easy or logic is difficult.
- d) That logic is not difficult or mathematics is not easy.
- e) That if not many students like logic, then either mathematics is
not easy or logic is not difficult.
Here's my solution (in detail). I'm not sure if I did all this correctly. I just want to make sure if I made any mistake!
$D(x):$ "$x$ is difficult"
$S(x):$ "Many students like $x$"
$l$ and $m$ to represent logic and mathematics.
Assumptions:
- $D(l) \lor \lnot S(l)$
- $\lnot D(m) \to \lnot D(l)$
Conclusions:
- a) $\; S(l) \to D(m) $
- b) $\; D(m) \to \lnot S(l) $
- c) $\; D(l) \lor D(m) $
- d) $\; \lnot D(l) \lor D(m) $
- e) $\; \lnot S(l) \to (D(m) \lor \lnot D(l)) $
I'm quite sure that the first conclusion (a) is valid.
$
1) D(l) \lor \lnot S(l)
$
: Assumption 1
$
2) S(l) \to D(l)
$
: Logical equivalence of (1)
$
3) \lnot D(m) \to \lnot D(l)
$
: Assumption 2
$
4) D(l) \to D(m)
$
: Contrapositive of (3)
$
5) S(l) \to D(m)
$
: Hypothetical syllogism on (2) and (4)
The second conclusion (b) is invalid because since the conclusion (a) is valid and this conclusion is
$
D(m) \to \lnot S(l) \equiv S(l) \to \lnot D(m) \not\equiv S(l) \to D(m)
$
The third conclusion (c) is invalid because the assumption (2) is not equivalent to this conclusion:
$
\lnot D(m) \to \lnot D(l) \equiv \lnot D(l) \lor D(m) \not\equiv D(l) \lor D(m)
$
The fourth conclusion (d) is valid because the assumption (2) is equivalent to this conclusion$
\lnot D(m) \to \lnot D(l) \equiv \lnot D(l) \lor D(m)
$
The last conclusion (e) is invalid followingly according to the rules of inference:
$
1) D(l) \lor \lnot S(l)
$ :Assumption 1
$
2) \lnot D(m) \to \lnot D(l) \equiv \lnot D(l) \lor D(m) \equiv \lnot D(l) \lor (\lnot D(l) \lor D(m))
$
$
3) \lnot S(l) \lor (\lnot D(l) \lor D(m))
$ : Resolution rule on (1) and (2)
$
4) S(l) \to (\lnot D(l) \lor D(m))
$ : Logical equivalence of (3)
$
5) S(l) \to (\lnot D(l) \lor D(m)) \not\equiv \lnot S(l) \to (\lnot D(l) \lor D(m))
$
Best Answer
Saying that a conclusion in propositional logic is valid means that it is tautologically true, which is not what you really mean. If the premises are true and the argument is valid, then you could simply call the conclusion “true” (instead of the stronger “valid”).
Your translations, and your proofs of (a) and (d), are all correct.
For ease of reading though, I am resymbolising the three atomic propositions:
$P: \quad D(l)$
$Q: \quad S(l)$
$R: \quad D(m)$
Assumptions:
(1) $P \lor \lnot Q$
(2) $\lnot R \to \lnot P$
Conclusions:
(a) $\; Q \to R $
(b) $\; R \to \lnot Q $
(c) $\; P \lor R $
(d) $\; \lnot P \lor R $
(e) $\; \lnot Q \to (R \lor \lnot P) $
Your disproof of (b) is invalid. Firstly, you've merely shown that the (derived) true statement (a) isn't logically equivalent to conclusion (b); this is a weaker assertion than that (a) doesn't entail (b). Besides, so what if (a) doesn't entail (b)? If, for example, premise (1) entails (b), then that's sufficient for conclusion (b) to be true, isnt it?
To disprove conclusion (b), just cite the truth assignment $(P,Q,R)=(T,T,T),$ which immediately has the premises both true yet conclusion (b) false.
Your argument in (c) is fallacious for the same reason.
Instead, the truth assignment $(P,Q,R)=(F,F,F)$ immediately disproves conclusion (c).
Your argument in (e) has 3 errors: firstly, as above, conclusion $Y$ can be a consequence of premise $X$ without being logically equivalent to $X;$ so the non-equivalence in line 5 does not disprove conclusion (e). Secondly, the equivalence in line 2 is false; but I think you meant implication instead, in which case it is true. Thirdly, line 3 is not a valid inference step.
In fact, conclusion (e) $$\lnot Q \to (R \lor \lnot P)$$ is true whenever the premises are true. Now, by the definition, the conditional $\to$ is true whenever its consequent (i.e., the part of the proposition after the arrow) is true, right? Well, according to part (d), whenever the premises are true, the consequent of the conditional statement (e) is true. Therefore, conclusion (e) is true.
Addendum (see comment below)