A few weeks back I made a comment on the ontological argument that generated some feed back. One of the things that I said is that the modal version of the ontological argument (MOA) depends on the modal logic of S5 and S5 is dubious. Hence, the MOA is dubious as well. And this comment generated a question and a complaint: What is S5? Prove that S5 is suspect.
To begin an answer we must construct a basic logic for handling the concepts of “possibility” and “necessity”. That is, we must construct a basic modal logic that we will call K (after Kripkie):
Guide to symbols: P • Q reads P and Q
P v Q reads P or Q
P ⊃ Q reads If P then Q
P ≡ Q reads P if and only if Q
-P reads It is not the case that P
cP reads It is necessary that P
⋄P reads It is possible that P
K = Normal Predicate Logic:
Code:
Rules of Inference:
P ⊃ Q plus P entails Q (Modus Pones)
P ⊃ Q plus –Q entails –P (Modus Tollens)
P ⊃ Q plus Q ⊃ R entails P ⊃ R (Hypothetical Syllogism)
P v Q plus –P entails Q (Disjunctive Syllogism)
P v Q plus P ⊃ R plus Q ⊃ R entails R (Simple Dilemma)
P v Q plus P ⊃ R plus Q ⊃ S entails R v S (Constructive Dilemma)
P • Q entails P (Simplification)
P plus Q entails P • Q (Conjunction)
P entails P v (anything) (Addition)
Rules of Replacement:
- (P • Q) is equivalent to -P v –Q DeMorgan’s Rule
- (P v Q) is equivalent to -P • -Q DeMorgan’s Rule
(P • Q) is equivalent to (Q • P) Commutativity
(P v Q) is equivalent to (Q v P) Commutativity
[P • (Q • R)] is equivalent to (P • Q) • R] Associativity
[P v (Q v R)] is equivalent to (P v Q) v R] Associativity
P is equivalent to - -P Double Negation
P ⊃ Q is equivalent to -Q ⊃ -P Transposition
P ⊃ Q is equivalent to -P v Q Material Implication
(P ≡ Q) is equivalent to (P ⊃ Q) • (Q ⊃ P) Mat. Equiv.
(P • Q) ⊃ R] is equivalent to [P ⊃ (Q ⊃ R)] Exportation
P • P is equivalent to P Tautology
P v P is equivalent to P Tautology
- (unary) operators for possibility and necessity
cP is equivalent to - ⋄ -P
⋄P is equivalent to -c-P
- The Distribution Axiom
c (P ⊃ Q) is equivalent to cP ⊃ c Q
But K doesn’t allow us to do (prove) things that we should be able to do (prove). So we strengthen K to S4:
S4 = K
Code:
+ Reflexivity
cP ⊃ P
+ Something Called 4
cP ⊃ c cP
But some want a modal system that allows them to do (prove) even more, S5. They construct S5 as follows:
S5 = K
Code:
+ Reflexivity
cP ⊃ P
+ Something called E
⋄P ⊃ c⋄P
and all of this seems very innocent. But S5 entails the dubious modal claim:
Code:
⋄cP ⊃ P = If it is possible that it is necessary that P then P.
Let us call the above claim “magic” because of its amazing powers. Here is the (reductio) proof that S5 entails magic.
- -(⋄cP ⊃ P) ACP: Assumption for conditional proof.
- ⋄cP • -P From 1 by MI and DM
- -P From 2 by SIMP
- ⋄-P ⊃ c⋄-P An instance of E.
- -P ⊃ ⋄-P
- -P ⊃ c⋄-P From 4 and 5 by HS
- c⋄-P From 3 and 6 by MP
- c-cP From 7 by definition
- -⋄cP From 8 by definition
- ⋄cP From 2 by SIMP
- -(⋄cP ⊃ P) ⊃ (⋄cP • -⋄cP) From 1 – 10, Conditional Proof
-
Code:
(⋄cP ⊃ P) From 11 and 12 by MT
So, S5 leads to magic, (⋄cP ⊃ P). And magic is the key to the MOA:
(1) ⋄c(God exist) ⊃ God Exist. fl an instantiation of magic
That is: If it is possible that it is necessary that God exist then God exist.
(2) ⋄c(God exist)
That is: It is possible that it is necessary that God exist.
(3) Therefore, God exist.
So, by a little bit of magic we have proved that God exists, or have we? First, think about what premise (2) is claiming: It is possible that necessarily God exists. But if ⋄X, doesn’t this suggest that ⋄-X In this case, isn’t it also possible that necessarily God doesn’t exist? Sure it is. And if it is, couldn’t we run the counter-MOA?:
(1) ⋄c(God doesn’t exist) ⊃ God doesn’t exist.
That is: If it is possible that it is necessary that God doesn’t exist then God doesn’t exist.
(2) ⋄c(God doesn’t exist)
That is, it is possible that it is necessary that God doesn’t exist.
(3) Therefore, God doesn’t exist.
