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9 January, 22:38

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

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  1. 9 January, 22:56
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    It is NOT TRUE

    Step-by-step explanation:

    ꓯyꓱxP (x, y)

    means that for each value of y there exist x such that P (x, y) is true

    whereas

    ꓱxꓯyP (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-example

    Consider

    P (x, y) the following proposition

    x-y = 0, for x, y integers.

    Given an integer y, there is another integer x (namely, x=-y) such that

    x-y = 0

    so, ꓯyꓱxP (x, y) is TRUE

    If 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 and

    x-1 = 0

    x-2 = 0

    and we conclude 1=2, which is a contradiction.

    So ꓱxꓯyP (x, y) is NOT TRUE (FALSE)
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