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Suppose there are two propositions $X$ and $Y$.

What is the opposite of the statement "$X$ and $Y$ is true"?

I am guessing it is that either $X$ or $Y$ or both of them are false. Is this correct? And if so how can we arrive at this solution from mathematical logic notation?

triomphe
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  • Not (X and Y) is true. Also, (not X) or (not Y) is true. – 2'5 9'2 Oct 14 '13 at 04:31
  • "Either $X$ or $Y$ or both of them" is what is usually meant when we say "$X$ or $Y$". The "both of them"-part is implicitly contained in the definition of "or". If you want to exclude the "both of them"-part, the logical term is "$X$ xor $Y$" (xor is short for "exclusive or"). – Arthur Oct 14 '13 at 13:58
  • Do you mean "X, and 'Y is true'", or "'X and Y' is true". In the latter case, the statement is just "X and Y"; there is no need to append "is true", because the claim that the statement is true is not part of the statement itself, rather the claim that the statement is true is a claim about the statement. – Carl Mummert Oct 15 '13 at 11:05

2 Answers2

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This is one of DeMorgan's Laws. You can prove it with a truth table easily. If you want to prove it from axioms, you need to specify the axioms and derivation rules available.

Ross Millikan
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The negation of $A \land B$ is given by $$\lnot (A \land B)$$ which is equivalent, by one of DeMorgan's Laws, to $$\lnot A \lor \lnot B$$

So yes, you are correct in your conjecture: If it is not the case that both $A$ and $B$ are true, this means that either $A$ is not true, or $B$ is not true, (or neither $A$ nor $B$ is true). But note that in logic and mathematics, the use of "or" is always taken to be the inclusive $\;(\lor)\;$ sense of the word "or", where $(P \;\text { or }\;Q) \equiv P\lor Q$ means exactly $P$ or $Q$ or both $P$ and $Q$, so the added clause "or both P and Q" is unnecessary.

amWhy
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  • Thanks you very much. I have a question though. In this formulation you don't take into account the part of my statement that is in quotation X and Y "is true". You only take the part "X and Y". So can't we write a logical equation like $X \land Y = true$ and take the negation of both sides. But that doesn't give the same result as then $\lnot X \land \lnot Y = false$ is satisfied only when both $X, Y$ are true. Which is equivalent to the original statement. Does that mean we disregard that part of the statement? How do we write the statement either X or Y is false?Please explain. many thanks. – triomphe Oct 14 '13 at 13:34
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    When we write the statement $X \land Y$, we are affirming the truth of ($X$ and $Y$). When we write the statement $\lnot (X \land Y)$, we are denying the truth of the statement $(X$ and $Y)$. (Hence, we are taking $(X$ and $Y)$ to be false.) And when we are given a premise $X \land Y$, we take as given the truth of $X \land Y$ in order to deduce what follows from the truth of $X \land Y$. – amWhy Oct 14 '13 at 13:40
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    Also, if you write, as an equation, $(X \land Y) = \text{ true}$, the negation of each side is $\lnot(X \land Y) = ;\text{false},;\iff \lnot X \lor \lnot Y = ;\text{false},;$ again, by DeMorgan's on the left-hand-side. – amWhy Oct 14 '13 at 13:46
  • Yea I made a mistake in the equation. Thanks. Even with your correction to the equation, it is satisfied only if both $X, Y$ are true. But that is not the opposite of the original statement $X$ and $Y$ is true. So does that means the original statement is not represented by the equation? Why is this inconsistency? – triomphe Oct 14 '13 at 13:52
  • You asked in your original question what the opposite of $X \land Y$ is. The opposite of $X \land Y$ is $\lnot (X\land Y) \equiv \lnot X \lor \lnot Y$, which is true if one or both of $X$, $Y$ is false. But if you negate both sides of the equation $X \land Y = true\tag{1}$, you assert precisely the equivalent statement $\lnot(X \land Y) = false$, which is not the opposite of $(1)$, it is equivalent to $(1)$. It's like the equation $x = 1$. The opposite of $x$ would be $x = -1$. But if we negate both sides of the equation $x = 1$, we get $-x = -1$, which is equivalent to $x = 1$. – amWhy Oct 14 '13 at 13:59
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    So what I'm saying is that the opposite of $X \land Y = true$ is actually $X \land Y = false$, which means $\lnot(X\land Y) = true$. – amWhy Oct 14 '13 at 14:04
  • Thanks you very much. Got it this time. – triomphe Oct 14 '13 at 14:08
  • You're welcome, MLT. ;-) – amWhy Oct 14 '13 at 14:10