This exercise uses the function Map Color and predicates In(x, y), Borders (x, y), and Country(x), whose arguments are geographical regions, along with constant symbols for various regions. In each of the following we give an English sentence and a number of can- didate logical expressions. For each of the logical expressions, state whether it (1) correctly expresses the English sentence; (2) is syntactically invalid and therefore meaningless; or (3) is syntactically valid but does not express the meaning of the English sentence. a. Paris and Marseilles are both in France. (i) In(Paris ^ Marseilles, France). (ii) In(Paris, France) 1 In(Marseilles, France). (iii) In(Paris, France) v In(Marseilles, France). b. There is a country that borders both Iraq and Pakistan. (i) 3c Country(c) ^ Border(c, Iraq) ^ Border(c, Pakistan). (ii) 3c Country(c) = (Border(c, Iraq) ^ Border(c, Pakistan)]. (iii) (c Country(c)] = [Border(c, Iraq) ^ Border(c, Pakistan) (iv) 3c Border(Country(c), Iraq ^ Pakistan). c. All countries that border Ecuador are in South America. (i) Vc Country(c) ^ Border(c, Ecuador) = In(c, South America). (ii) Vc Country(c) = (Border(c, Ecuador) → In(c, South America)]. (iii) Vc (Country(c) = Border(c, Ecuador)] = In(c, South America). (iv) Vc Country(c) ^ Border(c, Ecuador) A In(c, South America). d. No region in South America borders any region in Europe. (i) -[3 c, d In(c, South America) ^ In(d, Europe) 1 Borders (c,d)]. (ii) Vc,d (In(c, South America) ^ In(d, Europe)] = - Borders (c,d)]. (iii) Vc In(c, South America) = 3d In(d, Europe) 1 - Borders(c,d). (iv) Vc In(c, South America) → Vd In(d, Europe) → Borders(c,d). e. No two adjacent countries have the same map color. (i) Vx, y - Country(x) V-Country(y) V – Borders(x, y) V -(MapColor (x) = MapColor(y)). (ii) Vx,y (Country(x) ^ Country(y) ^ Borders (x, y) ^ -(x = y)) → -(MapColor(x) = Map Color(y)). (iii) Vx, y Country(x) ^ Country(y) ^ Borders(x, y) ^ -(MapColor(x) = Map Color(y)). (iv) Vx,y (Country(x) ^ Country(y) ^ Borders(x, y)) → MapColor(x y).