1
vote
Prove the formula $(\forall x Px \wedge \forall x Qx) \leftrightarrow \forall x [Px \wedge Qx]$
$\forall x \ Px \land \forall x \ Qx \implies \begin{cases} \forall x \ Px \implies Px \\ \forall x \ Qx \implies Qx \end{cases} \implies Px \land Qx \implies \forall x \ (Px \land Qx)$ (via ...
- 152
1
vote
Accepted
Do ∃x(Dog(x)) and ∃x(¬Dog(x)) contradict each other?
in the real world,
there exist objects which are dogs i.e. ∃x(Dog(x))
In other words: "something is a dog".
there exist objects which are not dogs i.e. ∃x(¬Dog(x))
In other words: "...
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