If I work, I cannot study.
Either I study or I pass mathematics.
I worked…
Therefore I passed mathematics.
I have translated this into
A : I work
B : I study
C : I pass mathematics
[A ^ B’ ] ^ [ B v C ] ^ [ A -> C ]
When you work this out, the arguement is invalid.
What I am wondering is, is the above symbolic logic statement a correct interpretation of the english, or should it read … [A ^ B’ ] ^ [ B v C ] -> [ A -> C ]
Note the " ->" “implies” symbol between the middle and the last section, as opposed to the " ^ " “and” symbol in my original version."
Thanks for your reply, but I’m only a first year student, and I don’t understand your answer.
The “either I study or I pass mathematics” schtick was given to us by the prof. We have only studied very simple logic. The prof advised us that he wanted us to break the English into a symbolic logic statement using “A”, “B” & “C”. I don’t understand what the " ~ " means, when you use it.
On preview, I see that Libertarian has the answer. But Libertarian, doesn’t A imply C - the word “therefore” seems to suggest that it does. Please advise. I agree that the statement is not a tautology.
The ~ means “not”. So A -> ~B means “If A then not B.”
You’ve established that A is true, so B is not true.
You’ve established that either B or C is true. B is not true, so C must be true.
Either way, really. If A implies C, then both A and C drop out immediately and the contradiction is even more apparent. The statements as they were given to you are a bit scrambled. The premise of the argument is A. So using Eris’s notation (which is a bit more common, ~ = not):
A
A -> ~B
C v B
Conclusion:
C
The last step is a disjunctive syllogism. The argument simply says that you passed mathematics without studying.
Not. The statement is, “If I work, [then] I cannot study.” You can look at it as an implication:
WORK implies NOT STUDYING, A -> ~B
or a conjunction,
NOT both WORK and STUDYING, ~(A & B)
which are logically equivalent.
The modus ponens is a rule of deduction from implication, which says that
( (P -> Q) & P ) -> Q,
if P implies Q, and P is true, then Q Q is true.
So we have, A -> ~B, if I work I cannot study, and A, I worked, hence ~B, I didn’t study.
Then we have the disjunction,
either I STUDY or I PASS MATH, B V C.
The syllogism runs
( (P V Q) & ~P ) -> Q.
Our form is
Either I study or I pass mathematics, B V C, but I didn’t study, ~B, so I passed mathematics, C.
The whole tautology is
( (A -> ~B) & (B V C) & A )-> C.
I’m using & for AND while I think your OP is using ^.
That depends on your teacher. There is no High Holy Notation. There are certain conventions, but lots of people wiggle a lot. And especially so since there is so much ASCII symbology these days. Eris has given you a comprehensive and correct analysis.
Yes, Achilles, (A -> B’) ^ (B V C) ^ A -> C is a tautology.
You can think of implication as defined in terms of negation and disjunction, so that (A -> B) is equivalent to (A’ V B). They have the same truth table semantics. Likewise, applying DeMorgan’s to the latter, you get (A ^ B’)’ (still using your notation).
Alternatively, you can think of negation as being defined in terms of implication and false, so that A’ is equivalent to (A -> False) (they have the same truth tables, again), and then disjunction is definable in terms of implication:
Some equivalences which help are:
“Not both P and [not Q]”, “P implies Q”:
~(P & ~Q) == P -> Q
“P or Q”, “[Not P] implies Q.”
P V Q == ~P -> Q
where I’m using == to mean “is identical to”.
If you run into trouble understanding a proof in one way, sometimes these transformations can help. I do better understanding with AND and OR rather than implication because almost all of my logic experience comes from digital circuits, and there is no “implies” chip (not that they couldn’t be constructed, of course, from P NAND NOTQ).
Also, note why the disjunctive syllogism works. It is the modus ponens. Saying, “Either P or Q; ~P, hence Q” is the same thing as saying, “~P implies Q; ~P, hence Q.” If you are more comfortable with disjunction (OR) than implication, learn the disjunctive syllogism and remember that you can turn implications into one. If you are more comfortable with the modus ponens, then learn that and know you can always transform a disjunctive syllogism into one.
As a complete aside, all the logical operations can be defined using only AND and NOT operations. If you’ve got those down, know that enough practice can get you anywhere.
If I work, I cannot study: [A -> B’]
If-then statements are always implications. p->q is equivalent (in first-order predicate calculus) to p’ v q, making this [A’ v B’]
Either I study or I pass mathematics: [B v C]
You’re good here.
I worked: [A]
Now, take the conjunction of all three:
A^[A’ v B’]^[B v C] ==
[A^A’ v A^B’]^[B v C] ==
[0 v A^B’]^[B v C] ==
A^B’^[B v C] ==
A^B’^B v A^B’^C ==
0 v A^B’^C ==
A^B’^C
Therefore you worked, didn’t study, and still passed mathematics. Try pulling it off on your qualifying exams, though…
