# Solve Hoare Method related problem

Exercise 1

{(pr = 0 => exists y(1 < y < i and x mod y = 0)) and (pr = 1 => forall y(1 < y < i => x mod y != 0) ) and i <= x && i < x && x mod i = 0}

pr := 0

i : = i+1

{(pr = 0 => exists y(1 < y < i and x mod y = 0)) and (pr = 1 => forall y(1 < y < i => x mod y != 0) ) and i <= x }

Exercise 2

{(pr = 0 => exists y(1 < y < i and x mod y = 0)) and (pr = 1 => forall y(1 < y < i => x mod y != 0) ) and i <= x && i < x && !(x mod i = 0)}

i:=i+1

{(pr = 0 => exists y(1 < y < i and x mod y = 0)) and (pr = 1 => forall y(1 < y < i => x mod y != 0) ) and i <= x }

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