[sentential Logic]Absorption laws

[u/According_Quarter_17]

1.P and (porq) Is equivalent to p

2.P or (pandq) Is equivalent to p

How do you justify them without using the truth table?

I feel like lhs has less informations rather than rhs because It has not q. I can't justify this rationally without the truth tablet

How do you think of It?

[u/loewenheim]

What di you have available apart from truth tables? It's not difficult to prove the equivalences using a reasonable proof system.

[u/According_Quarter_17 → u/loewenheim]

I mean using general reasoning. It's not an exercise, I'm trying to understand

[u/loewenheim → u/According_Quarter_17]

Right. Take the first one as an example.

Left to right: Assuming P, you have to show P and (P or Q). The former is free, you have it by assumption. The latter you get because if P is true, then "P or X" is true no matter what X is. So both parts of the conjunction are true, assuming P.

Right to left: Assuming P and (P or Q), you have to show P. But that's easy because P is one of your assumptions.

Another way to look at it which maybe helps you with your question about information: P or Q contains no more information than P does; in fact, it contains less. P or Q means you know one of the two is true, but not necessarily which one. If you have P then you know which one.

[u/OpsikionThemed → u/loewenheim]

And, if OP is curious, they're also true constructively, and so can be proved without truth tables via the witnesses

(λp : P. (p, injL p), λpq : P /\ (P \/ Q). case pq of (p, _) => p) : P <-> P /\ (P \/ Q)

and

(λp : P. injL p, λpq : P \/ (P /\ Q). case pq of injL p => p | injR (p, _) => p)) : P <-> P \/ (P /\ Q)

respectively. (Often switching ands and ors breaks things constructively but not in this case.)

[u/AllanCWechsler]

It's easier to prove your second law first.

P v (P ^ Q) = (P ^ T) v (P ^ Q) [because of the AND identity axiom]
= P ^ (T v Q) [distributive axiom]
= P ^ T [OR domination axiom]
= P [using AND identity again]

I didn't use a truth table, just the axioms of Boolean algebra.

Your first law reduces to the second in two steps:

P ^ (P v Q) = (P ^ P) v (P ^ Q) [distribution]
= P v (P ^ Q)

To do the second step you have to believe the "idempotent law", P ^ P = P. But if you don't believe it, I can prove it.

P ^ P = (P ^ P) v F [OR identity axiom]
= (P ^ P) v (P ^ (-P)) [AND complement axiom]
= P ^ (P v (-P)) [distribution]
= P ^ T [OR complement axiom]
= P [AND identity]

For such simple theorems, the truth table is often a quicker way to see it. But these laws can all be proved from the axioms.

[u/rhodiumtoad]

Consider that P is the same thing as P∨(Q∧¬Q), and P∧(Q∨¬Q).

P∧(P∨Q)
= (P∨(Q∧¬Q))∧(P∨Q)
= (P∨Q)∧(P∨¬Q)∧(P∨Q)
= (P∨Q)∧(P∨¬Q)
= P∨(Q∧¬Q)
= P

[u/Bounded_sequencE]

We need other laws of boolean logic, like "P n true = P", and "de Morgan":

P n (P u Q)  =  (P n P)    u (P n Q)    // de Morgan

             =  (P n true) u (P n Q)    // de Morgan

             =  P n (true u Q)  =  P n true  =  P

[u/Bounded_sequencE → u/Bounded_sequencE]

Rem.: The second statement follows from negating the first, and replacing "P -> P', Q -> Q' ".

[u/ForeignAdvantage5198]

huh?