To show that an argument is valid, why can we not assume that both its premises and conclusion are true then show that there's no contradiction?
Example:
If $x^2=4$ and $x\neq-2,$ then $x=2.$
Proof
Suppose that $x=2$ (assuming the conclusion).
Then $x^2=4$ and $x\neq-2.$
Thus, if $x^2=4$ and $x\neq-2$, then $x=2.$
With proof by contradiction, it is legitimate to assume that the conclusion is false; why?
Suppose that I promise that if you sweep my porch, then I will give you a reward, and furthermore I promise that the reward will not be an old pair of sneakers.
“Maybe the reward will be a bag of gold!” you say to yourself. “Suppose the reward is a bag of gold! MJD said he would give me a reward for sweeping his porch, but it would not be a pair of old sneakers. A bag of gold is a reward, and it is not an old pair of sneakers. MJD must be going to give me a bag of gold!”
Then you sweep my porch and I give you a pack of gum. “Here's your reward,” I say.
You wanted a bag of gold. “I proved logically that you were going to give me a bag of gold!” you complain.
“Well,” I reply, “it seems that your proof was wrong.”
I really laughed when reading MJD's excellent answer.
However you may want a more explanatory answer.
When you prove that "$A$ and $B$ are not in contradiction", what you really prove is that it cannot be proven that $(A \land B)$ is false (without any additional hypothesis).
Note that this does not prove that $(A \land B)$ is true: if $A$ or $B$ include some free variable $x$, and you have proven that it cannot be proven that $(\forall x, A \land B)$ is false, it may just mean that for some values of $x$, $(A \land B)$ is false, but it is true for other values of $x$.
If you have proven that $(\forall x, \text{it cannot be proven that } (A \land B) \text{ is false})$, then, assuming your theory is complete, you could deduce that $(\forall x, (A \land B) \text{ is true})$.
But anyway, in the example you present, you do not at all prove that the premises and conclusion are not contradictory. To prove that, you would need to prove that all possible logical reasonings from those hypotheses never reach a contradiction. There is an infinity of possible reasonings, so the arguments required are much, much more complex than what you wrote.
In proof by contradiction, the situation is different. We want to prove $A \Rightarrow B$. So we assume $\lnot B$, and we prove there is a contradiction with $A$. This is much easier than proving that there is no contradiction. You just have to find one contradiction, not prove that all reasonings never reach a contradiction.
Then, assuming the theory is consistent, this proves that $A \land \lnot B$ is false, so its contrary, $\lnot A \lor B$ is true. And this is logically equivalent to $A \Rightarrow B$.
(A∧B) is false". Do you mean instead that (A↔B) can be proven true? After all, 1+1=8 and 1+1=9 aren't in contradiction yet their conjunction is false.
– ryang
May 16 '23 at 03:10
X, what you really prove is that Y" sounds like you are saying that X implies Y, but in fact it is Y that implies X. However, my observation is premised on the idea that "P and Q contradict each other" precisely when they negate each other (since in logic, a contradiction is the conjunction of a sentence and its negation). $\quad$ You're now clarifying that that quotation is a definition; this is fine, but the way you wrote it doesn't read as a definition.
– ryang
May 16 '23 at 12:54
If $x^2=4$ and $x\neq-2,$ then $x=2.$
Proof
Suppose that $x=2$ (assuming the conclusion).
Then $x^2=4$ and $x\neq-2.$
Thus, if $x^2=4$ and $x\neq-2$, then $x=2.$
In your first two steps, you proved the given statement's converse $$C\implies A \land B.$$ Then I think you implicitly inferred that $$\color\red {C\implies} C \land A\land B,$$ though you didn't make it explicit. Then you implicitly inferred that $$\color\red {C\implies} \Big(A \land B\implies C\Big).\tag0$$ So far, so good. Unfortunately, you carelessly wrote statement $(0)$ just as $$A \land B\implies C,\tag3$$ making Step 3 an invalid inference! You'd neglected to discharge the supposition $\color\red C$ that was made in Step 1.
In other words, you have not validly deduced the required statement $(3);$ you have validly deduced statement $(0),$ which says that if $\color\red C$ (i.e., the supposition $\color\red {x=2}$ that you made in Step 1) is actually true then statement $(3)$ is true. Unfortunately, statement $(0)$ is not useful information, because it is a tautology: no knowledge of mathematics or elementary algebra is required to validly deduce statement $(0),$ which can be proven simply using a truth table.
Begging the question is the fallacy of arguing by assuming the conclusion. In your Step 1, assuming the consequent x = 2 of the given statement amounts to assuming that the statement is true (notice that if B is true then A→B is automatically true), that is, assuming the conclusion of the proof/argument; tautology $(0)$ exhibits the structure of this fallacy.
To show that an argument is valid, why can we not assume that both its premises and conclusion are true then show that there's no contradiction?
If the argument's premise $P$ and conclusion $C$ are both true, then by definition they don't contradict each other (i.e., $P\not\equiv\lnot C$).
When you say "show that there's no contradiction" (if you in fact mean "...leads to no contradiction" then how would you actually do that??), do you perhaps instead mean the weaker condition $\text“P_1,\ldots,P_n,C$ are consistent with one another”, which is an immediate consequence of $(P_1\land\ldots \land P_n \land C)$ being true? In other words, are you perhaps just meaning to ask this simpler question:
The answer is No: in real analysis,
has premise and conclusion both true yet is an invalid argument, because the corresponding conditional is not a tautology, because in complex analysis this conditional is actually false.
With proof by contradiction, it is legitimate to assume that the conclusion is false; why?
Because this assumption is merely provisional and is discharged by the end of the proof. On the other hand, your proof attempt does not eventually discharge the $\color\red{\text{assumption}}$ that $\color\red C$ is true. (The point is that when we make a supposition/assumption, we need to keep track of it and remember to also consider the case in which it is actually false.)
P.S. To say that the argument $(P_1\land\ldots \land P_n,\,\therefore C)$ is valid is to say that its premise forces the conclusion to be true even in an alien world/context; remember, we are investigating the logical truth of $(P_1\land\ldots \land P_n \to C)$ regardless of the truth value of "every squared number is nonnegative" in any particular world.
Why is that an invalid argument? If the elements are real numbers, then that should be valid?
"Definition. An argument is valid if and only if in every case where all the premises are true, the conclusion is true"
– enoopreuse22 May 17 '23 at 18:26John is my brother, therefore dogs bark corresponds to a true implication but not a valid argument/reasoning. In the bullet points, I indicate that the argument P, ∴C is valid precisely when P→C is a tautology; referring to your definition ("...in every 'case' where all the premises..."), P→C being a tautology means that as you vary the meanings of P and C (i.e., in every 'case'/context) P→C remains true. My postscript is explaining the same thing.
– ryang
May 18 '23 at 04:58
Also how could i have discharged the assumption that $C$ is true?
– enoopreuse22 May 18 '23 at 20:21I have to wonder about your thought process here coming up with this example. Why did you decide to assert that $x\neq -2$? I assume it's because if you said
$$\text{Suppose } x^2 = 4\text{. Then, } x = 2. $$
that you wouldn't be true. But that's exactly the problem with your "proof": it can also be used to prove false claims. After all, if we suppose $x=2$, we see that $x^2 = 2^2 = 4$. There is no hypothesis that rules out $2$, and so we must conclude that the statement is true. It's not though, because $x = -2$ is also not contradicted by the hypothesis.
To boil it down, your "proof" technique is "proof by converse".
To prove $p\implies q$, you show that $q\implies p$.
To prove that $(x^2=4 \wedge x\neq -2) \implies (x = 2)$, you show that $(x=2) \implies (x^2=4 \wedge x\neq -2)$.
If $p$ and $q$ are true, these statements are equivalent. But if $p$ is true and $q$ is false, $q\implies p$ is true while $p\implies q$ is false. This means that your technique will sometimes claim something is true when in reality it is false, thus not a proof. By contrast, the proof by contrapositive works because $p\implies q$ and $\neg q \implies \neg p$ are equivalent. This also implies that a "proof by inverse" would also not work, since it is equivalent to your "proof by converse".
