Lecture Notes, 13 Sep 2010
Last Update: 13 September 2010
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- Show that (p → (q → r))
((p ∧ q) → r):
To show that 2 propositions are logically equivalent,
we need to show that their truth tables have the same input-output (I-O)
T-table for (p → (q → r)):
||(q → r)
||(p → (q → r))|
| || || || || |
T-table for ((p ∧ q) → r):
||(p ∧ q)
||((p ∧ q) → r)|
| || || || || |
Note that the I-O columns are the same,
even though the intermediate columns are different.
- One topic of CSE 191 is learning how to do proofs. Here's
Let A, B be propositions.
Then A ≡ B iff (A ↔ B) is a tautology.
I.e., two propositions (A, B) are equivalent iff their biconditional (A ↔ B) is a tautology.
- Note that the book gives this as the definition of "logically
This kind of thing happens frequently in math:
definition is another's theorem
(and, whenever that's the case, the
first mathematician's theorem is the second mathematician's
- Note that there are 3 different uses of a kind of "biconditional" in this
- The "≡" says that two propositions are logically
- The "iff" says that two mathematical-English sentences
("A ≡ B" and "(A ↔ B)
is a tautology") are necessary and sufficient for each other.
- The "↔" connects A with B to form
a molecular proposition.
We have to show that a sentence of the form "P iff Q" is the case,
where P = "A ≡ B"
and where Q = "(A ↔ B) is a tautology".
To do that, we need to show two things:
- If P, then Q
- If Q, then P
Therefore, QED for the complete proof.
- To show: if P, then Q
we need to assume (or suppose, or make believe) that P is the case
and then show that—under that assumption (or under that
supposition, or in that hypothetical situation)—Q is the case
Suppose A ≡ B, & show (A ↔ B)
is a tautology:
Because A ≡ B, we can replace "≡" with its
For all rows of the t-table, tval(A) = tval(B)
But, if we now add a column for (A ↔ B) to the t-table,
it will get filled in with Ts, because of the semantics of ↔
So, for all rows, tval(A ↔ B) = T.
But this is the definition of a tautology!
So, (A ↔ B) is a tautology
which is what we wanted to prove ("QED") in this first part of the proof
[This is where we ended today, but, for the sake of completeness, here's
the rest of the proof, which we'll review in class on Wednesday.]
- To show: if Q, then P
we need to assume that Q, and then show that P:
Suppose (A ↔ B) is a tautology, & show A
Because (A ↔ B) is a tautology,
we can replace "is a tautology" by its definition:
for all rows, tval(A ↔ B) = T
But the only way for tval(A ↔ B) = T
tval(A) = tval(B), by the semantics of ↔
So, for all rows, tval(A) = tval(B)
But this is the definition of logical equivalence!
So, A ≡ B,
which is QED for the second part of the proof.
Text copyright © 2010 by William J. Rapaport
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