If you have been following the discussion in Plantinga on Free Will and Omiscience you will have seen that I have been struggling to construct a proof of (1), which says that if God knows that I will do some action before I actually do it then it is necessary that I actually do it, from (2), which says that it is necessary that if God knows what I will do some action in advance then I will actually do it.
(1) K(G,R,a) –> []D(R,a)
(2) [](K(G,R,a) –> D(R,a)
So far the two attempts that I have made have both been invalid because of some bonehead mistakes. This has been driving me crazy for the past couple of days, but now I think I got it, in fact it almost seems too simple (which probably means I made another bonehead mistake!)…
I thought that it would be easier if I did not include quantifiers, but I think that is what actually confused me. So, what I really want to prove is (1′), which says that for any action x if God knows that I will do it in advance then it is necessary that I actually do it, from (2′), which you can figure out for yourself.
(1′) (x)(K(G,R,x) –> [](D(R,x))
(2′) (x)[](K(G,R,x) –> D(R,x))
this actually turns out to be quite easy (I *think* 🙂 ).
1. ~(x)(K(G,R,x) –> []D(R,x)) assume as a theorem
2. (Ex)~(K(G,R,x) –> []D(R,x)) 1, by definition
3. (Ex)~~(K(G,R,x) & ~[]D(R,x)) 2, by def
4. (Ex) (K(G,R,x) & ~[]D(R,x)) 3, by def
5. K(G,R,a) & ~[]D(R,a) 4, EI
6. K(G,R,a) 5, CE
7. []K(G,R,a) 6, necessitation
8. ~[]D(R,a) 5, CE
9. (x)[] (K(G,R,x) –> D(R,x)) assumption (2′)
10. [](K(G,R,a) –> D(R,a)) 9, UI.
11. []K(G,R,a) –> []D(R,a) 10, distribution
12. ~[]D(R,a) –> ~[]K(G,R,a) 11, contraposition
13. ~[]K(G,R,a) 8,11 MP
14. []K(G,R,a) & ~[]K(G,R,a) 7,13 CI
15. (x)(K(G,R,x) –> [](D,R,x)) 1-14 reductio
What I didn’t notice before was that since we are assuming 1 as a theorem and we can get K(G,R,a) from that then we can use the rule of necessation, which says that if phi follows from a theorem then phi is necessary, to get []K(G,R,a).
So, free will is incompatible with God’s omniscience…