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Fact: There are two balls, one white and one black, in one bottle.

Define:

p : someone takes one ball from the bottle.
q : the ball is white 
r:  the ball is black

Now, (p →(q ∨ r)) → ((p →q)∨(p →r)) is a tautology; proof:

p → p
p → q = ¬ p V q

p →(q ∨ r) = ¬ p V q V r
(p →q)∨(p →r) = (¬ p V q) ∨(p →r) = (¬ p V q) ∨ (¬ p v r) = ¬ p V q V r

And this is a well-known inference rule:

A
A → B 
-----------
B

Consider my argument with the same form:

p →(q ∨ r)                            true
(p →(q ∨ r)) → ((p →q)∨(p →r))        tautologically true
----------------------------------------------
(p →q)∨(p →r)                         false, because (p →q) and (p →r) are both false

Why does my valid argument have a false conclusion even though its premises are all true?

EDIT

“someone takes one ball from the bottle” is not a proposition as it doesn’t have a truth value.

Aren't these propositions?

True: someone takes one ball from the bottle.   
False: someone does not take one ball from the bottle.

Otherwise, what are propositions?

ryang
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showkey
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    Your q and r are not well-formed because "the ball" is undefined. – Anne Bauval Aug 16 '24 at 07:15
  • As @AnneBauval said; If you fix one ball to solve this issue, then indeed that ball is white or it is black. – Maximilian Janisch Aug 16 '24 at 07:19
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    Firstly, “someone take one ball from the bottle” is not a proposition as it doesn’t have a truth value. You have to change it to something like “the ball is either white of black” to make this make sense… – David Gao Aug 16 '24 at 07:21
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    … But then there is an implicit universal quantifier in your proposition. What your premise actually is is that, “for any ball, if it is either white or black, then it is white or it is black” (which is true). Or, symbolically, $\forall x, p(x) \to (q(x) \vee r(x))$. While $(p \to (q \vee r)) \to ((p \to q) \vee (p \to r))$ is a tautology, your supposed conclusion is actually $(\forall x, p(x) \to q(x)) \vee (\forall x, p(x) \to r(x))$ (i.e., “for any ball, if it is either black or white then it is white; or, for any ball, if it is either black or white then it is black”, which, … – David Gao Aug 16 '24 at 07:23
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    … as you observed, is false), but $(\forall x, p(x) \to (q(x) \vee r(x))) \to (\forall x, p(x) \to q(x)) \vee (\forall x, p(x) \to r(x))$ is *not* a tautology, so no contradiction arises. – David Gao Aug 16 '24 at 07:25
  • To make these more like propositions, you could say "if someone has taken a ball from the bottle", "a white ball was removed from the bottle", and "a black ball was removed from the bottle." It is then true that $(p \to q) \vee (p \to r)$, but we can't say which one of the two implications is the true one. – David K Aug 16 '24 at 08:17
  • I suppose, if you insist, you can understand “someone take one ball from the bottle” as a proposition meaning “there is a person and a point in time such that the said person takes a ball from the bottle at that point in time”. Then it does have a truth value as that event either happened or didn’t. But that’s hardly what “someone take one ball from the bottle” means in the sentence “if someone take one ball from the bottle, then the ball is white or the ball is black”. (Or, at the very least, I wouldn’t understand the sentence that way.) The phrase “someone take one ball from the bottle”… – David Gao Aug 16 '24 at 08:20
  • … seems to me as indicating the setting of an event, rather than the premise of a material implication (after all, in English, “if…, then…” can mean lots of things, not necessarily material implication). A proposition, on the contrary, is supposed to have a well-defined truth value - if you speak of a proposition, someone should be able to reasonably ask you back whether it is true, and it should be the case that you can state either it is true or it is false. – David Gao Aug 16 '24 at 08:20
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    "because (p →q) and (p →r) are both false": why? –  Aug 16 '24 at 13:15

2 Answers2

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As said in the comments, there is a problem in your way of framing the sentences. As such they cannot be claimed to be propositions. It is like saying: $$2x<3$$ is a proposition whenever $x$ can take integer values. Unless and until you specify which $x$, you cannot decide the truth or falsity of this statement.

In the same fashion, say your set is $S=\{black, white\}$ and $x$ represents an element of this set, then your statement is like:

If Robert picks $x$ then $x$ is black.

However, the following turns out to be true:

If Robert picks $x$ then $x$ is either black or white

as there are only two choice - black or not black (meaning white).

Now what you are arguing is as follows:

If Robert picks $x$ then $x$ is white - False

If Robert picks $x$ then $x$ is black - False

We do not know the truth or falsity of the above two as they are not propositions.

Edit:

Moreover suppose the statement "If Robert picks an element of $S$ then the element which he picked is white" is false, then Robert should have picked black ball. Thus the statement $$\text{If Robert picked an element of $S$ then the element which he picked is black}$$ should be true.

Yathi
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Fact: There are two balls, one white and one black, in one bottle.

Define:

p : someone takes one ball from the bottle.
q : the ball is white 
r:  the ball is black

Argument:

p →(q ∨ r)                            true
(p →(q ∨ r)) → ((p →q)∨(p →r))        tautologically true
----------------------------------------------
(p →q)∨(p →r)                         false, because (p →q) and (p →r) are both false

Your intended argument is more naturally framed using quantifiers than in propositional logic. (See the final paragraph below.)

It is possible to contrive your intended argument in propositional logic, as long as you recognise that we are formulating synthetic implications (not analytic implications) based on a pre-defined context. In the following, I'm really just changing the tenses in your example; to make clearer the descriptive nature of our conditionals. The given context:

  • Bag X initially contained exactly two balls, one black and one white. A ball was drawn from some bag.
  • $P:\iff$ the ball had been drawn from Bag X
  • $Q:\iff$ the ball is white
  • $R:\iff$ the ball is black

Then this is indeed a valid argument (of the form Modus Ponens) whose premises and conclusion are all true:

 P → (Q ∨ R)
(P → (Q ∨ R))  →  ((P → Q) ∨ (P → R))
--------------------------------------
                   (P → Q) ∨ (P → R)

The above conclusion is false only for the assignment $(P,Q,R)=$ (true,false,false); since this assignment contradicts the given context, the above conclusion must actually be true.

If the white ball had been drawn, then the synthetic conditional P → Q is simply true.

Your main mistake was mistranslating the predicate-logic sentence "drawing a ball from Bag X implies that the ball is white" as P → Q instead of ∀b ( X(b) → W(b) ).

ryang
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