Let's compare a couple of arguments.
Logic is easy if it is fun, and it is fun, so it must be easy. F
-> E, F |- E
Logic is easy only if it is fun, and it is fun, so it must be easy.
E -> F, F |- E
in both cases we'll start off by listing the
variables alphabetically
| E | F | F->E | F | E |
| true | true | true | true | true |
| true | false | true | false | true |
| false | true | false | true | false |
| false | false | true | false | false |
| E | F | E->F | F | E |
| true | true | true | true | true |
| true | false | false | false | true |
| false | true | true | true | false |
| false | false | true | false | false |
In the second pattern we see something that should not happen when an argument is deductively valid. By definition, a valid argument rules out the possibiity of true premises and a false conclusion. We have a "bad line" that shows this can happen, and this is enough to tell us the pattern is invalid. (Imagine a story in which logic is very hard but it's fun anyway--like this course.)
Now let's compare the truth tables when we rewrite both arguments as
single expressions with implication signs as the main connectives (as though
we now stress the hypothetical quality of the premises). Also, instead
of "true" and "false" we will use the numbers 1 and 0.
| ((F->E) | & | F) | -> | E | ((E->F) | & | F) | -> | E |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 |
In the lefthand wff we have the value of the conditional as true all the way down, but that is not so in the righthand wff.
We could show the same thing with all valid argument forms. All of them are tautologies.
This fact allows us a very mechanical way of establishing validity for
both propositional and monadic predicate logic by means of truth tables.
(Later we will see why there is no mechanical test for predicate logic
when we move into two-place or dyadic predicates.)
But let's vary the story and say that all students are ambitious, and some ambitious individuals do work hard. Now we recognize that we could have Alberto be someone who is not a student even though he is ambitious. He might work hard, but there might not be any student who keeps up with him so that it would be imaginable that no student works hard.
(Ex(Sx&Ax)&Ax(Ax->Wx))->Ex(Sx&Wx) (Ax(Sx->Ax)&Ex(Ax&Wx))->Ex(Sx&Wx)
Here we could set up an interpretation that presents individuals a and b and considers all possibilities,but since this would be like having a truth table with six variables (and the number of lines doubles for each variable) it is not a particularly efficient method.
Instead we could show that in the righthand form there would be at least one exception so that it is not a tautology, but any intended exception in the lefthand form would lead to a contradiction.
Aa Ab Sa Sb Wa Wb
(Ex(Sx&Ax)&Ax(Ax->Wx))->Ex(Sx&Wx)
1 1
1 0 0
1
1 1 x
0
0
For this not to be a tautology, we have to
have one instance of the implication being false, and this could only happen
if the antecedent is true while the consequent is false. To have
the consequent false we'll say Alberto is a student but not working hard
(while someone else not a student--Barbara, for instance-- might be working
hard). However, now we make it impossible to say that all students
work hard.
Aa Ab Sa Sb Wa
Wb (Ax(Sx->Ax)&Ex(Ax&Wx))->Ex(Sx&Wx)
1 1
0 1 1
1 1
1 1
0
0
Alberto satisfies the condition that there
is someone who is ambitious and working hard, but we have the possibility
that no student is both ambitious and working hard. This shows us
that the argument form is not a tautology.