Two statements are logically equivalent if they have the same truth table inputs and outputs. How do I know if one statement can be inferred from another?

Does that just mean for inputs they share, they also share the same outputs?

Let me give an example of what I'm trying to do:

1.) If the car door is unlocked then Jimmy can enter the car

I understand that to mean P->Q

An equivalent statement to P->Q would be one with the same truth table (like ~P OR Q). They are equivalent because they have the same truth table:

P Q P->Q ~P OR Q





But I want to determine if another statement can be inferred from statement #1.

2.) If the car door is locked then Jimmy can not enter the car.

~P -> ~Q

Since ~P->~Q has a different truth table I know it's not equivalent and it can't be infered. But I could have a statement like '(P OR Q) AND (P->R)' and I want to know if (Q OR R) can be inferred from it.

Do I just compare the truth table values for Q and R?

  • $\begingroup$ Judging from the use of "truth table" I assume this is a question about propositional calculus; please let me know if it is not. $\endgroup$ – Asaf Karagila Feb 2 '13 at 16:08

If you think of logical equivalence as meaning $A \equiv B$, then this is also denoted $A \iff B$. So in the case of logical equivalence, $A \implies B$ AND $B \implies A$. Each side is implied by the other.

Here is the truth-table for Logical Equivalence ($A \equiv B$ or $A \iff B$:

enter image description here

In contrast: Logical inference is a one-sided implication: either $$A \rightarrow B\quad \text{or}\quad B \rightarrow A$$

$A \rightarrow B$ is defined to be equivalent to $\lnot A \lor B$. $A \rightarrow B$ is FALSE ONLY when $A$ is true, and $B$ is false.

Put differently, $A \rightarrow B$ is TRUE WHENEVER $A$ is false, OR whenever $B$ is true, or both.

Here's the truth table for logical inference (material implication):

enter image description here

So it is not at all equivalent to logical equivalence.


If you want to determine whether $(p \rightarrow q) \implies (\lnot p \rightarrow \lnot q)$, you can construct a truth-table to determine if this is true for all possible truth value assignments of p and q, and/or for which truth value assignments it is not true that the implication holds:

enter image description here

So the implication is not always true: In particular, when $p$ is false, and $q$ is true, we have that $p \rightarrow q$, but not $\lnot p\rightarrow \lnot q$. Hence one cannot infer $\lnot p\rightarrow \lnot q$ from $p \rightarrow q$.

  • $\begingroup$ I enjoy you use tables as above for most of logic shape question, Amy. I hope you are getting well my dear friend. :+) $\endgroup$ – Mikasa Oct 29 '13 at 8:46

To add to amWhy's response.

A. "Two statements are logically equivalent if they have the same truth table inputs and outputs." Do be careful, for that's not true in general. There are logically equivalent statements other than those which are equivalent in virtue of some truth-functional structure which can be picked up by doing a truth-table. (For example, there are some F's is equivalent to it is false that there are no $F$'s but doing a truth-table won't help to elucidate the logical connections here.)

B. As to the question whether, given '(P OR Q) AND (P->R)', '(Q OR R)' can be inferred from it, yes you can do a truth-table to determine that (assuming the connectives here are truth-functional). You work out the values of '(P OR Q) AND (P->R)' and then '(Q OR R)' on each of the eight possible assignments of values to the three propositional variables here. And then you need to ask: is there any way of making the premiss true and the conclusion false? That is, is there any line of the table on which the premiss gets the value T and the conclusion the value F? If there is, you know the inference is not (truth-functionally) valid. And if there isn't, the inference is valid.


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