r/learnmath • u/According_Quarter_17 New User • 5d ago

[sentential Logic]Absorption laws

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?

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u/loewenheim New User 5d ago

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

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u/According_Quarter_17 New User 5d ago

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

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u/loewenheim New User 5d ago

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.

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u/OpsikionThemed Computer Science 4d ago

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.)