Proof writing question

[proud thicket]

I have P∨Q, P->A and ¬Q and I want to proof A.
I divided the PvQ into cases.
Case 1: P, since P->A, so A.
Case 2: Q, but it contradicted with ¬Q, so this case is impossible.
Since case 1 alone already exhausted all possible cases, I concluded that A.
Is this proof fine?

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[versed dirge]

since you have (P or Q) and -Q, you must have P

cases are not necessary, show that P and apply modus ponens as you did

@proud thicket

[proud thicket]

okay got it

i thought when i see or i must use cases

sry for late reply

thank you so much

+close

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@proud thicket

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