logically equivalent

Printable View

March 17th 2010, 06:15 PM
questionboy

logically equivalent

can anyone explain to me how to prove that p → q and not q → not p are logically equivalent without truth tables
March 17th 2010, 08:19 PM
Soroban

Hello, questionboy!

Quote:

Prove that $p\to q$ and $\sim\!q \to \;\sim\!p$ are logically equivalent without truth tables

$p \to q \;\equiv\;\sim\!p \vee q$ . . I call it "Alternate Definition of Implication" (ADI).

Start with the right side:

. . $\begin{array}{ccccc} \sim\!q \to \:\sim\!p & \equiv & q \:\vee \sim\!p & &\text{ADI} \\ \\ & \equiv & \sim\!p \vee q && \text{Comm.} \\ \\ & \equiv & p \to q && \text{ADI} \end{array}$
March 18th 2010, 12:32 AM
Seppel

Hello!

Quote:

Originally Posted by Soroban

Hello, questionboy!

$p \to q \;\equiv\;\sim\!p \vee q$ . . I call it "Alternate Definition of Implication" (ADI).

Start with the right side:

. . $\begin{array}{ccccc} \sim\!q \to \:\sim\!p & \equiv & q \:\vee \sim\!p & &\text{ADI} \\ \\ & \equiv & \sim\!p \vee q && \text{Comm.} \\ \\ & \equiv & p \to q && \text{ADI} \end{array}$

You should also mention the use of Double Negation.

Best wishes,
Seppel
March 18th 2010, 05:55 AM
emakarov

This is fine, but expressing $p\to q$ as $\neg p\lor q$ misses an important subtlety. One can define $\neg p$ as $p\to\bot$ where $\bot$ denotes a contradiction (e.g., 0 = 1). Indeed, $p$ and $\neg p$ imply $\bot$ by Modus Ponens, and deriving $\bot$ from $p$ proves $\neg p$ by Deduction theorem (or implication introduction).

Then $\neg q\to\neg p$ becomes $(q\to\bot)\to(p\to\bot)$ . Now, it is surprising that this formula is implied by $p\to q$ without using the fact that $\bot$ is a contradiction. I.e.,

$(p\to q)\to\Big((q\to r)\to(p\to r)\Big)$

is the transitivity of $\to$ , and it is derivable for any $r$ using only basic rules about implication.

This suggests that $p\to q$ implies $\neg q\to\neg p$ in a much stronger sense than just in Boolean logic.