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# sole2 - ANALYSIS EXERCISE 2SOLUTIONS 1(a(x(Ac)c(x Ac(x A(x...

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ANALYSIS EXERCISE 2–SOLUTIONS 1. (a) ( x ( A c ) c ) ⇔ ¬ ( x A c ) ⇔ ¬¬ ( x A ) ( x A ). (b) ( x ( A B ) c ) ⇔ ¬ ( x ( A B )) ⇔ ¬ ( x A ) ( x B ) [ ¬ ( x A ) ∧ ¬ ( x B )] [( x A c ) ( x B c )] ( x A c B c ) . (c) ( x ( A B ) c ) ⇔ ¬ ( x ( A B )) ⇔ ¬ ( x A ) ( x B ) [ ¬ ( x A ) ∨ ¬ ( x B )] [( x A c ) ( x B c )] ( x A c B c ) . 2. (a) ( x A 4 B ) ⇔ { [( x A ) ∧ ¬ ( x B )] [( x B ) ∧ ¬ ( x A )] } ⇔ { [( x A ) ( x B )] [( x A ) ∨ ¬ ( x A )] [ ¬ ( x B ) ( x B )] [ ¬ ( x B ) ∨ ¬ ( x A )] } ⇔ { [( x A ) ( x B )] [ ¬ ( x B ) ∨ ¬ ( x A )] } ⇔ { [( x A ) ( x B )] ∧ ¬ [( x B ) ( x A )] } ⇔ { ( x A B ) ∧ ¬ ( x A B ) } . (b) From (a) it follows that (0.1) ( A 4 B = ) [( A - B = ) ( B - A = )] . Now let us prove that

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sole2 - ANALYSIS EXERCISE 2SOLUTIONS 1(a(x(Ac)c(x Ac(x A(x...

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