Let P(x,y) be a propositional function if ꓯyꓱxP(x,y) is true does it necessarily follow that ꓱxꓯyP(x,y) is true? Justify your answer or give a counter-example

Accepted Solution

Answer:It is NOT TRUEStep-by-step explanation:ꓯyꓱxP(x,y)  means that for each value of  y there exist x such that P(x,y) is truewhereasꓱxꓯyP(x,y)means that there exists x such that for each value y, P(x,y) is true.In the second case, the same x must work for every element y.Counter-exampleConsider  P(x,y) the following propositionx-y = 0, for x, y integers.Given an integer y, there is another integer x (namely, x=-y) such that x-y = 0so, ꓯyꓱxP(x,y) is TRUEIf ꓱxꓯyP(x,y) were TRUE, then would exist a single unique value of x such that P(x,y) is TRUE for every integer y.Then P(x,1) and P(x,2) would be both TRUE andx-1 = 0x-2 = 0and we conclude 1=2, which is a contradiction.So  ꓱxꓯyP(x,y) is NOT TRUE (FALSE)