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) columns:T-table for (

*p*→ (*q*→*r*)):*p**q**r*( *q*→*r*)( *p*→ (*q*→*r*))T T T T T T T F F F T F T T T T F F T T F T T T T F T F F T F F T T T F F F T T T-table for ((

*p*∧*q*) →*r*):*p**q**r*( *p*∧*q*)(( *p*∧*q*) →*r*)T T T T T T T F T F T F T F T T F F F T F T T F T F T F F T F F T F T F F F F T 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
one:
**Theorem:**-
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.**Remarks:**- Note that the book gives this as the
*definition*of "logically equivalent".

This kind of thing happens frequently in math:- One mathematician's
definition is another's theorem

(and, whenever that's the case, the first mathematician's*theorem*is the second mathematician's*definition*)! - Note that there are 3 different uses of a kind of "biconditional" in this
theorem:
- The "≡" says that two propositions are logically equivalent.
- 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.

**proof:**-
We have to show that a sentence of the form "P iff Q" is the case,
- If P, then Q
- If Q, then P
- 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 caseSuppose

*A*≡*B*, & show (*A*↔*B*) is a tautology:-
Because

*A*≡*B*, we can replace "≡" with its definition:-
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*≡*B*:-
Because (

*A*↔*B*) is a tautology, we can replace "is a tautology" by its definition:- for all rows, tval(

*A*↔*B*) = TBut the only way for tval(

*A*↔*B*) = T is for 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.

where P = "*A*≡*B*"

and where Q = "(*A*↔*B*) is a tautology".To do that, we need to show two things:

- Note that the book gives this as the

Text copyright © 2010 by William J. Rapaport (rapaport@buffalo.edu)

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http://www.cse.buffalo.edu/~rapaport/191/F10/lecturenotes-20100913.html-20100913-2