Question

is it possible to prove \[p \vee (p \wedge q) \equiv p\]
without using truth tables?

Answer 1

A truth table would be a good way to prove this thing and it'd be a complete proof.

Answer 2

truth tables are too boring. it can't be solved by manipulation?

Answer 3

I've never heard of other proofs better than truth tables in logic... some exceptions are there though.

Answer 4

manipulation is better to prove this.. \[(p\vee q) \wedge (\neg p \vee r) \rightarrow (q\vee r) \equiv T\]

Answer 5

Like the statement \(\rm p \wedge q\Leftrightarrow q\wedge p\) doesn't need any proof because it's an axiom, but a truth table is still a good way to prove.

Answer 6

since truth tables will give 8 rows there

Answer 7

Uh-huh, yeah, manipulation is cool sometimes.

Answer 8

too bad it uses algebra (and math)

Answer 9

anyway...back to the main question...

Answer 10

how to prove that without t.t.'s?

Answer 11

Then, as you said, manipulation?

Answer 12

\[ \large p\vee(p\wedge q)\equiv(p\wedge\mathbb{T})\vee(p\wedge q) \]
\[ \large \equiv p\wedge(\mathbb{T}\vee q)\equiv p\wedge\mathbb{T}\equiv p \]

Answer 13

yes

Answer 14

what happened there?

Answer 15

I still love the traditional way :D

Answer 16

I'd rather stick with basics...
p∨(p∧q) -> (p∨p)∧(p∨q) by distribution
-> p ∧ (p∨q) since p∨p is just p
-> p by simplification: A ∧ B --> A
THUS, p∨(p∧q) -> p, by a series of hypothetical syllogisms

p -> p∨q, by addition; A --> A∨B
-> p ∧ (p∨q), conjunction
-> (p∨p) ∧ (p∨q), since p is the same as p∨p
-> p∨(p∧q), un-distributing p
THUS, p -> p∨(p∧q), by a series of hypothetical syllogisms

THEREFORE, p is EQUIVALENT to p∨(p∧q)

Answer 17

-> p by simplification: A ∧ B --> A

what??

Answer 18

If A and B is true, then A is true and B is true. Then A is true.

Answer 19

so...you're using t.t.'s

Answer 20

These are like axioms, which can be only be proven with truth tables. There are such things. Like proving A -> ~~A

Answer 21

actually...what you did was not an axiom....what you're comparing it to...is an axiom

Answer 22

Well, then I'm missing something. If you could enlighten me with its proof, then I'd be smarter than I am now... :)

Answer 23

we wouldn't want that to happen..now would we..

Answer 24

I certainly would...

Answer 25

aha, here we go:
http://en.wikipedia.org/wiki/Simplification
immediate inference

Answer 26

they are not axioms.

they are logical laws because they are tautologies.

axioms in mathematical logic are very different.

Answer 27

@helder_edwin duly noted
I will stop referring to them as such :)

Answer 28

your proof, @zugzwang, is very neat, though lengthy. also it requires less, "knowledge" than mine.