Logic: Lecture 3, 11 August 2015
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Derivability
Assume X, prove a
Formulas in X are concrete, not axiom schemes
X |- a
Redefine deriviations:
A sequence where each formula is an instance of an
axiom. obtained using MP or a member of X
Deduction theorem: X,a |- b iff X |- a -> b
Prove by inducion
Examples
1. |- a -> (b -> c) -> (b -> (a -> c))
a -> (b -> c) |- a -> (b -> c) (Membership)
a -> (b -> c),a |- b -> c (Deduction Theorem)
a -> (b -> c),a,b |- c (Deduction Theorem)
a -> (b -> c),b |- a -> c (Deduction Theorem)
a -> (b -> c) |- b -> (a -> c) (Deduction Theorem)
|- a -> (b -> c) -> (b -> (a -> c)) (Deduction Theorem)
2. |- ~~b -> b
1. |- (~b -> ~~b) -> ((~b -> ~b) -> b) (A3)
2. |- (~b -> ~~b) -> ((~b -> ~b) -> b) ->
(~b -> ~b) -> ((~b -> ~~b) -> b) (Example 1 above)
3. |- (~b -> ~b) -> ((~b -> ~~b) -> b) (MP, 1,2)
4. |- (~b -> ~b) (Instance of a -> a)
5. |- (~b -> ~~b) -> b (MP, 3,4)
6. |- ~~b -> (~b -> ~~b) (A1)
7. ~~b |- ~b -> ~~b (Deduction Theorem)
8. ~~b |- (~b -> ~~b) -> b (From 5)
9. ~~b |- b (MP, 7,8)
10. |- ~~b -> b (Deduction Theorem)
Completeness
Prove contrapositive: not derivable implies not valid
Consistency: not (|- ~a)
Claim:
a or b is consistent if either a is consistent or b is consistent
a and b is consistent if both a and b are consistent (Converse fails!)
Henkin-style proof of completeness:
Show that every consistent formula is satisfiable
Assuming this, completeness follows:
Suppose b is not derivable.
|- ~~b -> b, so ~~b is also not derivable (else, by MP, |- b)
So ~b is consistent
So ~b is satisfiable
So b is not valid.
Maximal consistent sets
Finite set is consistent if conjunction is consistent
Infinite set is consistent if all finite subsets are consistent
MCS is a consistent set that cannot be extended
Lindenbaum Lemma
Every consistent set can be extended to an MCS
Properties of MCS
a in X iff ~a not in X
a or b in X iff a in X or b in X
MCS X defines a valuation v_X = all p in X
Extends to all formulas: v_X |= A iff A in X
Now, Henkin:
a consistent => extend {a} to MCS X => v_X |= a => a is satisfiable
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