Anyone good at logic? Premise 1: ( f ^ e ) ^ q -> p Premise

>>173642
>( f ^ e ) ^ q -> p
>dingus puts parentheses around the wrong thing

is it
( f ^ e ^ q ) -> p
f ^ e ^ ( p -> q )

>>173642
You could use the 0, 1 table.

Suppose that both q and e are true. Then the Premise 1 implication holds since f^e^q is true hence p.
If either q or e is not true then the other is true if we want Premise 3 to hold. But then in Premise 1 f^e^q is not true, but from false assumption we can prove anything hence p is true as well in this case.
f must be true for Premise 2 to hold, and q (and) e can't be false at the same time since we have Premise 3.

You could also use a proof by contradiction.

Counterexample.

f = true
q = true
e = false
p = false

All the premises hold but the conclusion's false.

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>>173664
As far as I understand, an argument is invalid if a situation in which all premises are true, but the conclusion false, is possible.
(the critical row on truth tables)

So, you take P to be false (it's the false conclusion we're trying to prove)

Since f is a premise we take f to be true

Take both q and e as true (premise 3)

But this makes premise 1 false, because
x -> y returns false where x is true and y is false.

Is this argument thereby valid?

>>173685
It's q or e, not q and e. You can set one to false.

>>173687
No you can't, because that fucks up premise 1 (all ands), no?

>>173689
oh fuck no, because f -> f returns true

>>173690
You got it. That only returns false if it's F -> T

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>>173694
thanks anon