Logical Equivalences: prove p v ( ~p ^ (q v p)) = p ^ q?

Question

Key:
~ not
^ and
v or
How do you prove this to be true? I keep getting p v q

Answer 1

Try drawing out a truth table, and showing all possible truth combinations of p and q.

My recommendation is put in as many columns as needed. For example, obviously, you need a column each for p and q.

Then I recommend the following additional columns:
~p
q v p
~p ^ (q v p)
p v (~p ^ (q v p))
p ^ q

Then the equivalence is shown by the columns representing both sides of the equivalence agree with each other.

For more information on truth tables, try http://en.wikipedia.org/wiki/Truth_Table

Answer 2

P,

or not:

Not equals “negative”.

Negative p “and” {plus} q OR {plus} p.

Negative p cancels the p in the “or” statement. Negative p and p equals zero.

Take the parenthetic (q or p) and combine it with negative p, and you are left with q.

You now have p or q

P v Q

Answer 3

one million. (P > R) & (Q > R) Premise 2. (~P v R) & (Q > R) one million cloth Implication 3. (~P v R) & (~Q v R) 2 cloth Implication 4. (R v ~P) & (~Q v R) 3 Commutation 5. (R v ~P) & (R v ~Q) 4 Commutation 6. R v (~P & ~Q) 5 Distribution 7. (~P & ~Q) v R 6 Commutation 8. ~(P v Q) v R 7 De Morgan's regulation 9. (P v Q) > R 8 cloth Implication one million. (P v Q) > R Premise 2. ~(P v Q) v R one million cloth Implication 3. (~P & ~Q) v R 2 De Morgan's regulation 4. R v (~P & ~Q) 3 Commutation 5. (R v ~P) & (R v ~Q) 4 Distribution 6. (~P v R) & (R v ~Q) 5 Commutation 7. (P > R) & (R v ~Q) 6 cloth Implication 8. (P > R) & (~Q v R) 7 Commutation 9. (P > R) & (Q > R) 8 cloth Implication

Answer 4

p v ( ~p ^ (q v p))
Distributive property:
= p v ( (~p ^ q) v (~p ^ p) )
AND of disjoint statements = false:
= p v ((~p ^ q) v F)
Distributive property again:
= (p v ~p) ^ (p v q)
OR of disjoint statements = true:
= T ^ (p v q)
= p v q

I think you're right...good job.

Answer 5

= means equivalence

p v ( ~p ^ (q v p))

= Distributing p,

= (p v ~p) ^ (p v (q v p))

Commutation:
= (p v (q v p)) ^ (p v ~p)

Simplification:
= p v (q v p)

Commutation:
= p v (p v q)

Association:
= (p v p) v q

Since p v p = p (tautology)
= p v q


:D

Answer 6

p ∨ (¬p ∧ (q ∨ p))
(p ∨ ¬p) ∧ (p ∨ (q ∨ p))
T ∧ (p ∨ q)
p ∨ q

Your solution is correct. If you don't trust the above, just look at the truth table:

p -- q -- (q∨p) -- (¬p ∧ (q ∨ p)) -- p∨(¬p∧(q∨p))
T .. T ...... T ................ F .......................... T
T .. F ...... T ................ F .......................... T
F .. T ...... T ................ T .......................... T
F .. F ...... F ................ F ......................... F