Answer
a. True. Circle b is black and squares h and j are black also.
b. Formal version: ∃x(Circle(x) ∧ (∃y(Square(y) ∧ SameColor(x, y))))
c. Formal negation: ∀x(∼Circle(x) ∨ (∀y(∼Square(y) ∨∼SameColor(x, y))))
Work Step by Step
c. ~(∃x(Circle(x) ∧ (∃y(Square(y) ∧ SameColor(x, y)))))
$\equiv$ $\forall$x ~(Circle(x) ∧ (∃y(Square(y) ∧ SameColor(x, y))))
(by the law of negating an $\exists$ statement)
$\equiv$ $\forall$x (~Circle(x) $\lor$ ~(∃y(Square(y) ∧ SameColor(x, y))))
(by De Morgan's Law)
$\equiv$ $\forall$x (~Circle(x) $\lor$ ($\forall$y ~(Square(y) ∧ SameColor(x, y))))
(by the law of negating an $\exists$ statement)
$\equiv$ $\forall$x (~Circle(x) $\lor$ ($\forall$y (~Square(y) $\lor$ ~SameColor(x, y))))
(by De Morgan's law)
