Make a table for such that each column represents a value:
(p <-> q) -> p
T T T T T
T
F
T
1 3 2 4 1
The last row represents the order in which the values are assigned.
When you complete the table, you could post it for verification.
Construct a truth table for the statement.
(p ↔ q) → p
8 answers
T T T T T
T F F T T
F T T F F
T T T F F
??
T F F T T
F T T F F
T T T F F
??
There are some corrections to make, starting from p (first column) and q (third column). Sorry that the first column was wrong, it should have read:
(p <-> q) -> p
T T T T T
T ? F
F ? T
F ? F
1 3 2 4 1
We usually give
p [TTFF] and
q [TFTF] in order to have all possible combinations.
The "<->" operator is true when both p and q are true, or both are false,
and "m->n" is false if and only if m is true and n is false.
(p <-> q) -> p
T T T T T
T ? F
F ? T
F ? F
1 3 2 4 1
We usually give
p [TTFF] and
q [TFTF] in order to have all possible combinations.
The "<->" operator is true when both p and q are true, or both are false,
and "m->n" is false if and only if m is true and n is false.
i am still so confused.... :(
Work with a simple table:
The biconditional p<->q is true if and only if both p and q are true, or both p and q are false.
p q p<->q
T T T
T F F
F T F
F F T
Now from this truth table, you will make another one for the conditional which is false if and only if p is true and q is false.
p q p->q
T T T
T F F
F T T
F F T
Now combine the two results to find
(p <-> q) -> p,
i.e. replace (p <-> q) with the results from the first table, and p remains [TTFF].
Figure out the result of the conditional according to the second table.
(p<->q) p (p<->q)->p
T T ?
F T ?
F F ?
T F ?
The biconditional p<->q is true if and only if both p and q are true, or both p and q are false.
p q p<->q
T T T
T F F
F T F
F F T
Now from this truth table, you will make another one for the conditional which is false if and only if p is true and q is false.
p q p->q
T T T
T F F
F T T
F F T
Now combine the two results to find
(p <-> q) -> p,
i.e. replace (p <-> q) with the results from the first table, and p remains [TTFF].
Figure out the result of the conditional according to the second table.
(p<->q) p (p<->q)->p
T T ?
F T ?
F F ?
T F ?
Jen,
If you are still confused at this point, it may do you good to back up and study the truth tables for the basic logic operators:
∧, ∨ → ↔ ¬.
Once you have mastered them, the truth tables of compound propositions would be much easier to do.
Remember that you will soon be working with simplification of compound propositions by a dozen of known identities, which require the thorough knowledge of the simple logic operators.
For the current problem, you will need the following:
p q p<->q
T T T
T F F
F T F
F F T
p q p->q
T T T
T F F
F T T
F F T
If you are still confused at this point, it may do you good to back up and study the truth tables for the basic logic operators:
∧, ∨ → ↔ ¬.
Once you have mastered them, the truth tables of compound propositions would be much easier to do.
Remember that you will soon be working with simplification of compound propositions by a dozen of known identities, which require the thorough knowledge of the simple logic operators.
For the current problem, you will need the following:
p q p<->q
T T T
T F F
F T F
F F T
p q p->q
T T T
T F F
F T T
F F T
p q p <-> q (p<->q)->p
T T T T
T F F T
F T F T
F F T F
??
T T T T
T F F T
F T F T
F F T F
??
Excellent, all correct!
1 answer