without the consideration of , we have the following conclusions: xS.pxS.p xS.pxS.p We used "pp" during the proof to get the above conclusions, so they are not totally correct anymore with the consideration of . Let's create some conclusions without replacing p by p. Remind that, the following definitions about satisfaction of quantified predicates are still correct (I rephrased them here): xS.p means there exists S, it is the case that [x]p. xS.p means for every value S, we have [x]p. answer this question: substitute instead of type1 and type2 as described so that each of the following sentences is correct. Fill type 1 with word "some", "every" or "this". Fill type 2 with ""or "". xS.p means for type 1 state and for type 1 S, it is the case that [x] type 2 p. xS.p means for type 1 state and for type 1 S, it is the case that [x] type 2 p. xS.p means for type 1 state and for type 1 S, it is the case that [x] type 2 p. xS.p means for type 1 state and for type 1 S, it is the case that [x] type 2 p