# The Dilemma Once More

P1. If it is possible that necessarily there is an omniscient, omnipotent, omnibenevolent being, necessarily there is an omniscient, omnipotent, omnibenevolent being. (From axiom 5 of S5)

P2. Either the proposition “necessarily there is an omniscient, omnipotent, omnibenevolent being” entails the proposition “there is gratuitous evil and suffering” or it is not the case the proposition “necessarily there is an omniscient, omnipotent, omnibenevolent being” entails the proposition “there is gratuitous evil and suffering”. (From the Law of the Excluded Middle)

P3. For all propositions p if there is some proposition q such that it is not the case that p entails q, then possibly p. (Contraposition of the Principle of Explosion)

C1. If it is not the case the proposition “necessarily there is an omniscient, omnipotent, omnibenevolent being” entails the proposition “there is gratuitous evil and suffering”, it is possible that necessarily there is an omniscient, omnipotent, omnibenevolent being. [From P3]

C2. If it is not the case the proposition “necessarily there is an omniscient, omnipotent, omnibenevolent being” entails the proposition “there is gratuitous evil and suffering”, necessarily there is an omniscient, omnipotent, omnibenevolent being. [From P1 and C1, Hypothetical Syllogism]

P4. If the proposition “necessarily there is an omniscient, omnipotent, omnibenevolent being” entails the proposition “there is gratuitous evil and suffering”, gratuitous evil and suffering is not counter-evidence to the proposition “necessarily there is an omniscient, omnipotent, omnibenevolent being”.

C3. Either necessarily there is an omniscient, omnipotent, omnibenevolent being, or gratuitous evil and suffering is not counter-evidence to the proposition “necessarily there is an omniscient, omnipotent, omnibenevolent being.” (From P2,C2,P4 Constructive Dilemma)

 The axiom in S5 can be found here: https://en.m.wikipedia.org/wiki/S5_(modal_logic). So, given the axiom 5 of S5: ♢p → ☐♢p

Here is the proof for P1:

Let

Kx ≝ x is omniscient
Px ≝ x is omnipotent
Bx ≝ x is omnibenevolent

1 ~ ☐(∃x)[(Kx ∧ Px) ∧ Bx] (Assump. CP)
2 ~ ☐~~(∃x)[(Kx ∧ Px) ∧ Bx] (1 DN)
3 ♢~(∃x)[(Kx ∧ Px) ∧ Bx] (2 ME)
4 ☐♢~(∃x)[(Kx ∧ Px) ∧ Bx] (3 Axiom 5)
5 ☐~~♢~(∃x)[(Kx ∧ Px) ∧ Bx] (4 DN)
6 ☐~☐(∃x)[(Kx ∧ Px) ∧ Bx] (5 ME)
7 ~☐(∃x)[(Kx ∧ Px) ∧ Bx] → ☐~☐(∃x)[(Kx ∧ Px) ∧ Bx] (CP 1-6)
8 ~☐~☐(∃x)[(Kx ∧ Px) ∧ Bx] → ~~☐(∃x)[(Kx ∧ Px) ∧ Bx] (7 Contra)
9 ~☐~☐(∃x)[(Kx ∧ Px) ∧ Bx] → ☐(∃x)[(Kx ∧ Px) ∧ Bx] (8 DN)
10 ♢☐(∃x)[(Kx ∧ Px) ∧ Bx] → ☐(∃x)[(Kx ∧ Px) ∧ Bx] (9 ME)

 The Law of the Excluded Middle can be found here: https://en.m.wikipedia.org/wiki/Law_of_excluded_middle

 Contraposition can be found here: https://en.m.wikipedia.org/wiki/Contraposition

 The Principle of Explosion can be found here: https://en.m.wikipedia.org/wiki/Principle_of_explosion

Here is the proof that P3 is the contrapositive of the Principle of Explosion, which we will state as follows: (∀p)[~♢p → (∀q)(p ⊨ q)], for all propositions p, if p is impossible, then for all propositions q1, p entails q.

1 (∀p)[~♢p → (∀q)(p ⊨ q)] (Principle of Explosion)
2 ~♢φ → (∀q)(φ ⊨ q) (1 UI)
3 ~(∀q)(φ ⊨ q) → ~~♢φ (2 Contra)
4 (∃q)~(φ ⊨ q) → ~~♢φ (3 QN)
5 (∃q)~(φ ⊨ q) → ♢φ (4 DN)
6 (∀p)(∃q)~(p ⊨ q) → ♢p] (5 UG)

 Here is the proof that C1 follows from P3:

Let

G ≝ ☐(∃x)[(Kx ∧ Px) ∧ Bx]
E ≝ ‘there is gratuitous evil and suffering’

1 (∀p)(∃q)~(p ⊨ q) → ♢p] (P3)
2 ~(G ⊨ E) (Assump. CP)
3 (∃q)~(G ⊨ q) → ♢G (1 UI)
4 (∃q)~(G ⊨ q) (2 EG)
5 ♢G (3,4 MP)
6 ~(G ⊨ E) → ♢G (205 CP)
7 ~(G ⊨ E) → ♢☐(∃x)[(Kx ∧ Px) ∧ Bx] (6 def. of ‘G’)

Thus Line 7 (C1) follows from Line 1 (P3), QED.

 Hypothetical Syllogism can be found here: https://en.m.wikipedia.org/wiki/Hypothetical_syllogism

 This premise is defended on given a Bayesian interpretation of counter-evidence:
(∀p)(∀q){[P(p|q)<P(p)] ⊃ Cqp} (read as: for all proposition p and q, if the probability of q given p is less than the probability of q unconditioned, then q is counter-evidence for p).

If we assume G ⊨ E, then by Logical Consequence P(E|G) = 1, but if E is counter-evidence to G, then it must be the case that P(G|E) < P(G). But both of these statements about probabilities cannot be true.

According to Bayes’ Theorem:

P(E|G) = [P(E)/P(G)] x P(G|E)

So given P(E|G) = 1

We can infer:

P(G)/P(G|E) = P(E)

But given 0 ≤ P(E) ≤ 1, it is not possible for P(G)/P(G|E) = P(E) and P(G|E) < P(G), as whenever the denominator is less than the numerator, the result is greater than 1.

Hence, we must reject the assumption that [P(E|G) = 1] ∧ [P(G|E) < P(G)]. This provides us with the following defense of P4:

1 ~{[P(E|G) = 1] ∧ [P(G|E) < P(G)]} (Result from the proof by contradiction above)
2 ~[P(E|G) = 1] ∨ ~[P(G|E) < P(G)] (1 DeM)
3 [P(E|G) = 1] → ~[P(G|E) < P(G)] (2 Impl)
4 [G ⊨ E] → [P(E|G) = 1] (by Logical Consequence)
5 [G ⊨ E] → ~[P(G|E) < P(G)] (3,4 HS)

And line 5 is just what is meant by P4.

 Constructive Dilemma can be found here: https://en.m.wikipedia.org/wiki/Constructive_dilemma

 The proof of the entire argument is as follows:

1 ♢☐(∃x)[(Kx ∧ Px) ∧ Bx] → ☐(∃x)[(Kx ∧ Px) ∧ Bx] (Premise)
2 (G ⊨ E) ∨ ~(G ⊨ E) (Premise)
3 (∀p)(∃q)~(p ⊨ q) → ♢p] (Premise)
4 [G ⊨ E] → ~[P(G|E) < P(G)] (Premise)
5 ~(G ⊨ E) (Assump CP)
6 (∃q)~(G ⊨ q) → ♢G (3 UI)
7 (∃q)~(G ⊨ q) (5 EG)
8 ♢G (6,7 MP)
9 ~(G ⊨ E) → ♢G (5-8 CP)
10 ~(G ⊨ E) → ♢☐(∃x)[(Kx ∧ Px) ∧ Bx] (9 definition of ‘G’)
11 ~(G ⊨ E) → ☐(∃x)[(Kx ∧ Px) ∧ Bx] (1,10 HS)
12 ☐(∃x)[(Kx ∧ Px) ∧ Bx] ∨ ~[P(G|E) < P(G)] (2,4,11 CD)

Posted on May 14, 2017, in Arguments for God, Atheistic Arguments and tagged , , , , , . Bookmark the permalink. Leave a comment.