Trying to negate a long logic expression, I encountered problems concerning operator precedence. As I was trying to google this I was only ever finding programming precedences and boolean precedences, but not for mathematical statements. Consider this:
$$ \neg (( A \implies B ) \lor (C) ) $$
How I would solve this is to first:
$$ \neg ( A \implies B ) \wedge \neg (C) $$
and then
$$ (A \wedge \neg B ) \wedge \neg (C) $$
This seems like the most logical order to me. Could you please confirm if this is true and maybe provide some additional resources on operator precedence?
The real problem is a lot more convoluted and I would like to know if my method to work from the "outer" layer to the "inner" layer is correct. Outer and inner layer shall her refer to the brackets.
Edit:
If we replace $$(A \implies B )$$ with $$ (\forall , y \in M : P(x) \wedge Q(y) \implies P(y) )$$ we yield:
$$ (\forall , y \in M : P(x) \wedge Q(y) \implies P(y) ) \lor (C) $$
Trying to negate this new statement $$ \neg (\forall , y \in M : P(x) \wedge Q(y) \implies P(y) ) \lor (C) $$
in the same manner I get:
$$ (\forall , y \in M : P(x) \wedge Q(y) \wedge \neg P(y) ) \wedge \neg (C) $$
Sorry, I could not properly format my comment, therefore I'm commenting in response to Mr. Scott in this answer.
Are these two statements equivalent?
$$ (\forall , y \in M : P(x) \wedge Q(y) \implies P(y) )$$ $$ \forall , y \in M ( P(x) \wedge Q(y) \implies P(y) ) $$