Hilbert systems:
Sufficient to have 2 axioms and a rule:
Rules will be given, don’t need to be memorized.
Admissible rule:
Proof system K is sound and complete with respect to all frames, ⊢K φ iff ⊨ φ.
Soundness and completeness results (note: the frame classes need to be memorized)
Give derivation in K of ◇ φ ∧ □ (φ → ψ) → ◇ ψ. Same as example in book MLOM page 52.
First, work backwards from the goal towards an axiom or tautology:
◇ φ ∧ □ (φ → ψ) → ◇ ψ ≡ □ (φ → ψ) → (◇ φ → ◇ ψ) [you can rewrite a conjunction as an implication] ≡ □ (φ → ψ) → (¬ □ ¬ φ → ¬ □ ¬ ψ) [rewrite diamond to ¬ □ ¬] ≡ □ (φ → ψ) → (□ ¬ ψ → □ ¬ φ) [rewrite contrapositive (¬ a → ¬ b) to (b → a)] ≡ □ (φ → ψ) → □ (¬ ψ → ¬ φ) [box distribution over implication] ≡ □ (φ → ψ) → □ (φ → ψ) [again contrapositive] ≡ (φ → ψ) → (φ → ψ) [because if derivable (a → b), then derivable (□ a → □ b)]
We arrive at a tautology.
Then you write it out in a Hilbert-style proof, starting with the tautology/axiom. PROP means rewriting in propositional logic.
You can safely rewrite a conjunction to an implication: (a ∧ b → c) ≡ a → (b → c). Remember that implication is right-associative!
If you don’t trust me, I didn’t trust myself either so I made a truth table:
a | b | c | b → c | a ∧ b | a → c | a ∧ b → c | b → (a → c) | a → (b → c) |
---|---|---|---|---|---|---|---|---|
0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
You see that the right three columns all have the same values, so semantically the formulas are the same.