The truth-function conditional, also called the material conditional, is a binary boolean truth function whose values are as you described in the truth table; it is true for all inputs except when the antecedent (left hand side) is true and the consequent (right hand side) is false. The conditional is written with a variety of symbols, e.g., $P \to Q$, $P \supset Q$, and $P \Rightarrow Q$.
There is an inference rule called modus ponens which says that from a conditional $P \to Q$ and $P$, infer $Q$. This can be written in a number of ways, e.g.,
- $P$
- $P \to Q$
- $Q$ by modus ponens from 1 and 2.
or
$$ \begin{array}{c} P \quad P \to Q \\ \hline Q \end{array} $$
Because modus ponens is so important, and because in some axiomatic systems it is the only inference rule, it is sometimes called “the inference rule.” Most of the time, that usage should probably be avoided, because in practice (i.e., outside of those specific axiomatic systems), there are plenty of other inference rules, and modus ponens is just one among many.
Asserting that the conditional $P \to Q$ is true is simply to assert that either:
- a. $P$ is true and $Q$ is true; or
- b. $P$ is false and $Q$ is true; or
- c. $P$ is false and $Q$ is false.
Using the inference rule modus ponens lets us affirm that $Q$ is true, based on the prior assertion that $P \to Q$ is true and that $P$ is true.
Modus ponens is a sound inference rule because whenever all the premises are actually true, then the conclusion is also actually true. To see why this is the case, consider the premises to modus ponens. These are a conditional $P \to Q$ and $P$. If $P \to Q$ is true, then one of the three cases (a, b, c) described above must also be true. $P \to Q$ being true does not, by itself, ensure that $Q$ is true, because there is one case (c) in which $P \to Q$ is true, but $Q$ is false. However, with the additional requirement that $P$ is true, we are restricted to the case (a) in which $Q$ is also true. Thus, if the premises to modus ponens are true, then so is its conclusion.
Now, it's worth considering how this applies to the example that you gave. The example of “all men are mortal, Socrates is man, therefore Socrates is mortal”. It uses first-order reasoning, and is not actually a case of modus ponens, neither premise is a conditional. However, the proof does use _modus ponens_, in that it requires us to make the following inference.
- (Premise) If Socrates is a man then Socrates is mortal.
- (Premise) Socrates is a man.
- (Conclusion) Therefore, Socrates is mortal.
There are two important concepts to consider: soundness and validity. These mean slightly different things for inference rules and for arguments.
An inference rule is sound if whenever its premises are true, then its conclusion is true. As we saw earlier, modus ponens is a sound rule of inference. The term valid is not used concerning inference rules. We do not say that an inference rule is valid or invalid.
An argument is valid if each reasoning step is an application of a sound inference rule. This means that each sentence in the argument must be true if the earlier sentences that it is based on are true. It does not make the claim that those earlier sentence are true, but just that if they are true, then the current sentence is true. An argument is sound if its premises are, in fact, true. If an argument is both sound and valid, then its conclusion must be true.
So, both
- (Premise) If Socrates is a man then Socrates is mortal.
- (Premise) Socrates is a man.
- (Conclusion) Therefore, Socrates is mortal.
and
- (Premise) If Italy is a man then Italy is mortal.
- (Premise) Italy is a man.
- (Conclusion) Therefore, Italy is mortal.
are valid arguments, because they use only valid inference rules (namely, modus ponens). The first argument is sound because both of its premises are true. The second argument is unsound because one of its premises its second premise, “Italy is man,” is not true.
If the conclusion of an argument is not true, it means that the argument is either invalid or unsound. (Of course, it could also be both.)
The trick to this is to be very familiar with these laws and inference rules your professor has given you. You need to be able to recognize which rule to apply on the spot and this kind of logical thinking will take lots of practice, especially because there are a lot of laws, but getting the hang of these basic logic skills is crucial to understanding more complicated mathematical proofs later on. Proofs later on will often skip these logical steps and will use all of these rules without even naming them because if they did that, then the proofs would be unnecessarily long. Therefore, you have to be able to manipulate these laws and inference rules very quickly to be able to follow the reasoning of a mathematical proof.
By practicing manipulating these rules for yourself, you'll eventually just get an intuition for following how these complicated logical expressions have certain conclusions by looking at the expressions and manipulating them in your head. It might take a while, but once you get the hang of it, you won't need to look at all of these laws because you'll just be able to see which logical laws apply in a certain situation. I know that when I was learning these rules for the first time, they were very confusing and intimidating, much more than truth tables, but now, I use them all of the time and I think using them is often faster than using truth tables.
However, before you have this intuition on which law to apply, just go through the list and try applying different laws. Try to see which laws work and which laws don't in certain situations. By experimenting with these laws with many, many practice problems, you'll be able to gain an intuition for which laws should be used in which situations.
Problem a:
We have $P$ and $P \implies Q$. Therefore, by Modus Ponens, we know that $Q$.
By Double Negation, we know that from $Q$, we can deduce $\sim \sim Q$.
We have $\sim \sim Q$ and $\sim R \implies \sim Q$. Thus, by Modus Tollens, we have $\sim \sim R$.
By Double Negation, we know that from $\sim \sim R$, we have $R$, concluding the proof.
Problem b:
We have $\sim R \wedge Q$. By Simplification, we thus have $\sim R$.
By Addition, this means we have $\sim R \vee \sim Q$.
By De Morgan's Law, $\sim R \vee \sim Q$ is the same as $\sim (R \wedge Q)$.
By Commutative, this is the same as $\sim (Q \wedge R)$.
Thus, we have $P \implies (Q \wedge R)$ and $\sim (Q \wedge R)$. Therefore, by Modus Tollens, we have $\sim P$, concluding the proof.
Best Answer
A set has no truth value: it is a (mathematical) object.
Syntactically, $A \cap B$ is a term, i.e. a "name" of a set (names have reference but not truth value).
A statement about sets (in the context of a specified theory about sets) has a truth value.
$A \cap B \subseteq A$ is a statement asserting that "the set resulting from the intersection of two sets $A$ and $B$ is a subset of set $A$".
The statement to which we have to apply the rules of inference is "$x \in A \cap B$" (it is a statement because has the form: "name-Relation-name", like e.g. "John is son of Peter").
The proof works this way: "$x$ is element of $A \cap B$" is our starting point.
By definition of "intersection of sets" ($\cap$) we have that: "$x$ is element of $A$ and $x$ is element of $B$".
Using Simplification we get:
In symbols:
1) $x \in A \cap B$ --- premise: $x$ is an element whatever
2) $(x \in A) \land (x \in B)$ --- from 1) unwinding the definition of set intersection ($\cap$)
3) $x \in A$ --- from 2) by Simplification
4) $(x \in A \cap B) \to (x \in A)$ --- from 1) and 4) by Conditional Proof
5) $\forall x \ [(x \in A \cap B) \to (x \in A)]$ --- from 4) by Universal Generalization
What happens if $A$ is the emptyset ?
The proof still holds, in a simplified form.
We can jump directly to step 4): if $A= \emptyset$, then $x \in (A \cap B)$ is False and thus the conditional $(x \in A \cap B) \to (x \in A)$ is True (because: $\text F \to \text F$ is $\text T$).
Thus, either $A$ is empty, and thus proceed as above, or $A$ is not empty, and thus pick an element $x$ whatever in it (for sure there is one) and use the full proof above.