Relación 1

header {* 1ª relación de ejercicios *}

theory Relacion_1
imports Main 
begin

text {*
  --------------------------------------------------------------------- 
  Demostrar los siguientes lemas usando sólo las reglas básicas de
  deducción natural de la lógica proposicional.
  --------------------------------------------------------------------- 
*}

lemma I: "A ⟶ A"
by (rule imp_refl)

-- "Se puede hacer automáticamente"
lemma I_d2: "A ⟶  A"
by auto

-- "Otra forma (más explicito)
lemma I: "A ⟶ A"
proof (rule impI)
  assume 1: "A"
qed




lemma 
  assumes 1:"A ∧ B" 
  shows  "B ∧ A" 
proof -
  from 1 have 2:"A" by (rule conjunct1)
  from 1 have 3:"B" by (rule conjunct2)
  from 3 2 show "B ∧ A" by (rule conjI)
qed




lemma "A ∧ B ⟶ A ∨ B"
proof
  assume "A ∧ B"
  hence "A" ..
  thus "A ∨ B" ..
qed




lemma
  assumes 1:"(A ∨ B) ∨ C"
  shows "A ∨ ( B ∨ C)"
proof -
note 1
 moreover
 { assume 2: "C"
   from 2 have 3:"B ∨ C" by (rule disjI2) 
   from 3 have "A ∨ (B ∨ C)" by (rule disjI2)}
 moreover
 { assume 4: "A ∨ B"
   from 4 have "(A ∨ B) ∨ C" by (rule disjI1) }
ultimately show "A ∨ (B ∨ C) " by (simp)
qed

-- Otra forma (más explícita)
lemma "(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)"
proof
  assume 1: "(A ∨ B) ∨ C"
  show "A ∨ (B ∨ C)"
  proof -
    note 1
    moreover{
      assume 2: "A ∨ B"
      have "A ∨ (B ∨ C)"
        proof -
          note 2
          moreover{
            assume 3: "A"
            from 3 have 4:"A ∨ (B ∨ C)" by (rule disjI1)
          }
          moreover{
            assume 5:"B"
            from 5 have 6: "B ∨ C" by (rule disjI1)
            from 6 have 7: "A ∨ (B ∨ C)" by (rule disjI2)
          }
          ultimately show "A ∨ (B ∨ C)" by (rule disjE)
        qed
      }
      moreover{
        assume 8: "C"
        from 8 have 9: "B ∨ C" by (rule disjI2)
        from 9 have "A ∨ (B ∨ C)" by (rule disjI2)
      }
      ultimately show "A ∨ (B ∨ C)" by (rule disjE)
    qed
qed




lemma Aux:
  assumes "A"
  shows "B ⟶ A"
using assms by simp

lemma K: "A ⟶ (B ⟶ A)"
proof
  assume 1: "A"
  from 1 show "B ⟶ A" by (rule Aux)
qed




lemma "(A ⟶ (B ⟶C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))"
proof
  assume 1:"A ⟶ (B ⟶ C)"
  show "(A ⟶ B) ⟶ (A ⟶C)"
    proof
      assume 2:"(A ⟶ B)"
      show "A ⟶C"
      proof
        assume 3:"A"
        from 2 3 have 4:"B" by (rule mp)
        from 1 3 have 5:"B ⟶ C" by (rule mp)
        from 5 4 show "C" by (rule mp)
      qed
    qed
qed

lemma "(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))"
proof
  assume 1:"(A ⟶ B)"
  show "((B ⟶ C) ⟶ (A ⟶ C))"
  proof 
    assume 2:"(B ⟶ C)"
    show "(A ⟶ C)"
    proof
      assume 3:"A"
      from 1 3 have 4:"B" by (rule mp)
      from 2 4 show "C" by (rule mp)
    qed
  qed
qed

lemma "¬¬A ⟶ A"
proof
  assume 1:"¬¬A"
  from 1 show "A" by (rule notnotD)
qed

lemma "A ⟶ ¬¬A"
proof
  assume 1:"A"
  from 1 show "¬¬A" by (rule contrapos_pn)
qed

lemma MT:
  assumes 1:"F ⟶ G" and 2:"¬G"
  shows "¬F"
proof (rule notI)
  assume 3:"F"
  from 1 and 3 have 4:"G" by (rule mp)
  from 2 and 4 show False by (rule notE)
qed

lemma "(¬A ⟶ B) ⟶ (¬B ⟶ A)"
proof
  assume 1:"(¬A ⟶ B)"
  show "(¬B ⟶ A)"
  proof
    assume 2:"¬B"
    from 1 2 have 3:"¬¬A" by (rule MT)
    from 3 show "A" by (rule notnotD)
  qed
qed

lemma "((A ⟶ B) ⟶ A) ⟶ A"
oops

lemma "A ∨ ¬A"
oops

lemma "(¬(A ∧ B)) = (¬A ∨ ¬B)"
oops

end