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      1 +++
      2 title = 'Lecture 8'
      3 +++
      4 # Lecture 8
      5 ## Modal tautologies
      6 
      7 ![Modal tautologies](modal-tautologies.png)
      8 
      9 ## Proof systems
     10 Hilbert systems:
     11 - proof is sequence of numbered formulas
     12 - every formula is: an axiom, or result of applying a derivation rule
     13 
     14 Sufficient to have 2 axioms and a rule:
     15 - K: φ → ψ → φ
     16 - S: (φ → ψ → θ) → (φ → ψ) → (φ → θ)
     17 - modus ponens: if φ and (φ → ψ), then ψ
     18 
     19 Rules will be given, don't need to be memorized.
     20 
     21 Admissible rule:
     22 
     23 ![Admissible rule definition](admissible-rule.png)
     24 
     25 Proof system K is sound and complete with respect to all frames, ⊢K φ iff ⊨ φ.
     26 
     27 <!-- TODO: add this to anki -->
     28 Need to memorize soundness and completeness results
     29 - K sound and complete for all frames
     30 - T sound and complete for all _reflexive_ frames
     31     - T: K with □ p → p
     32 - S4 sound and complete for all _reflexive-transitive_ frames
     33     - S4: T with □ p → □ □ p
     34 - S5 sound and complete for all frames with R an equivalence relation
     35     - S5: S4 with ¬ □ p → □ ¬ □ p
     36 
     37 ## Example of derivation
     38 Give derivation in K of ◇ φ ∧ □ (φ → ψ) → ◇ ψ.
     39 Same as example in book <abbr title='Modal Logic for Open Minds (Benthem)'>MLOM</abbr> page 52.
     40 
     41 First, work backwards from the goal towards an axiom or tautology:
     42 
     43 ```
     44 ◇ φ ∧ □ (φ → ψ) → ◇ ψ ≡ □ (φ → ψ) → (◇ φ → ◇ ψ)                 [you can rewrite a conjunction as an implication]
     45                       ≡ □ (φ → ψ) → (¬ □ ¬ φ → ¬ □ ¬ ψ)         [rewrite diamond to ¬ □ ¬]
     46                       ≡ □ (φ → ψ) → (□ ¬ ψ → □ ¬ φ)             [rewrite contrapositive (¬ a → ¬ b) to (b → a)]
     47                       ≡ □ (φ → ψ) → □ (¬ ψ → ¬ φ)               [box distribution over implication]
     48                       ≡ □ (φ → ψ) → □ (φ → ψ)                   [again contrapositive]
     49                       ≡ (φ → ψ) → (φ → ψ)                       [because if derivable (a → b), then derivable (□ a → □ b)]
     50 ```
     51 
     52 We arrive at a tautology.
     53 
     54 Then you write it out in a Hilbert-style proof, starting with the tautology/axiom.
     55 PROP means rewriting in propositional logic.
     56 
     57 1. (φ → ψ) → (¬ ψ → ¬ φ). PROP.
     58 2. □ (φ → ψ) → □ (¬ ψ → ¬ φ). DISTR, 1.
     59 3. □ (p → q) → □ p → □ q. modal distribution (i.e., this is an axiom in K that we use)
     60 4. □ (¬ ψ → ¬ φ) → □ ¬ ψ → □ ¬ φ. substitution, 3 (i.e., substitute stuff in the axiom).
     61 5. □ (φ → ψ) → □ ¬ ψ → □ ¬ φ. PROP, 2, 4.
     62 6. □ (φ → ψ) → ¬ ◇ ψ → ¬ ◇ φ. definition of ◇, □.
     63 7. □ (φ → ψ) → ◇ φ → ◇ ψ. PROP, 6.
     64 8. ◇ φ ∧ □ (φ → ψ) → ◇ ψ. PROP, 7.
     65 
     66 <details>
     67 <summary>Why you can rewrite a conjunction as an implication</summary>
     68 
     69 You can safely rewrite a conjunction to an implication: (a ∧ b → c) ≡ a → (b → c).
     70 Remember that implication is right-associative!
     71 
     72 If you don't trust me, I didn't trust myself either so I made a truth table:
     73 
     74 <table>
     75 <thead>
     76 <tr>
     77 <th>a</th>
     78 <th>b</th>
     79 <th>c</th>
     80 <th>b → c</th>
     81 <th>a ∧ b</th>
     82 <th>a → c</th>
     83 <th>a ∧ b → c</th>
     84 <th>b → (a → c)</th>
     85 <th>a → (b → c)</th>
     86 </tr>
     87 </thead>
     88 <tbody>
     89 <tr>
     90 <td>0</td>
     91 <td>0</td>
     92 <td>0</td>
     93 <td>1</td>
     94 <td>0</td>
     95 <td>1</td>
     96 <td>1</td>
     97 <td>1</td>
     98 <td>1</td>
     99 </tr>
    100 
    101 <tr>
    102 <td>0</td>
    103 <td>0</td>
    104 <td>1</td>
    105 <td>1</td>
    106 <td>0</td>
    107 <td>1</td>
    108 <td>1</td>
    109 <td>1</td>
    110 <td>1</td>
    111 </tr>
    112 
    113 <tr>
    114 <td>0</td>
    115 <td>1</td>
    116 <td>0</td>
    117 <td>0</td>
    118 <td>0</td>
    119 <td>1</td>
    120 <td>1</td>
    121 <td>1</td>
    122 <td>1</td>
    123 </tr>
    124 
    125 <tr>
    126 <td>0</td>
    127 <td>1</td>
    128 <td>1</td>
    129 <td>1</td>
    130 <td>0</td>
    131 <td>1</td>
    132 <td>1</td>
    133 <td>1</td>
    134 <td>1</td>
    135 </tr>
    136 
    137 <tr>
    138 <td>1</td>
    139 <td>0</td>
    140 <td>0</td>
    141 <td>1</td>
    142 <td>0</td>
    143 <td>0</td>
    144 <td>1</td>
    145 <td>1</td>
    146 <td>1</td>
    147 </tr>
    148 
    149 <tr>
    150 <td>1</td>
    151 <td>0</td>
    152 <td>1</td>
    153 <td>1</td>
    154 <td>0</td>
    155 <td>1</td>
    156 <td>1</td>
    157 <td>1</td>
    158 <td>1</td>
    159 </tr>
    160 
    161 <tr>
    162 <td>1</td>
    163 <td>1</td>
    164 <td>0</td>
    165 <td>0</td>
    166 <td>1</td>
    167 <td>0</td>
    168 <td>0</td>
    169 <td>0</td>
    170 <td>0</td>
    171 </tr>
    172 
    173 <tr>
    174 <td>1</td>
    175 <td>1</td>
    176 <td>1</td>
    177 <td>1</td>
    178 <td>1</td>
    179 <td>1</td>
    180 <td>1</td>
    181 <td>1</td>
    182 <td>1</td>
    183 </tr>
    184 </tbody>
    185 </table>
    186 
    187 You see that the right three columns all have the same values, so semantically the formulas are the same.
    188 
    189 </details>
    190