Razonamiento automático (2010-11) Razonamiento automático (2010-11)

RA_Relacion_1

theory Relacion_1
imports Main 
begin

(* Demostrar los siguientes lemas usando sólo las reglas básicas de deducción
   natural de la lógica proposicional. *)

lemma I: "A ⟶ A"
proof 
  assume 1: "A"
qed
  
lemma 0: "A ∧ B ⟶ B ∧ A"
proof 
  assume 1: "A ∧ B"
  from 1 have 2: "A" by (rule conjunct1) 
  from 1 have 3: "B" by (rule conjunct2)
  from 3 and 2 show "B ∧ A" by (rule conjI)
qed

lemma "A ∧ B ⟶ A ∨ B"
proof
  assume 1: "A ∧ B"
  from 1 have 2: "A" by (rule conjunct1)
  from 1 have 3: "B" by (rule conjunct2)
  from 2 show "A ∨ B" by (rule disjI1)
qed

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 S: "(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 and 3 have 4: "B" by (rule mp)
      from 1 and 3 have 5: "B ⟶ C" by (rule mp)
      from 5 and 4 show 6: "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 and 3 have 4: "B" by (rule mp)
      from 2 and 4 show 5: "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 and 2 have 3:"¬¬A" by (rule MT)
    from 3 show "A" by (rule notnotD)
  qed
qed

lemma "((A ⟶ B) ⟶ A) ⟶ A"
proof
  assume 1: "(A ⟶ B) ⟶ A"
  have 8: "¬¬A"
  proof (rule notI)
      assume 2: "¬A"
      have 3: "A ⟶ B"
      proof (rule impI)
        assume 4: "A"
        from 2 and 4 show "B" by (rule notE)
      qed
      from 1 and 3 have 5: "A" by (rule mp)
      from 2 and 5 show False by (rule notE)
  qed
  from 8 show "A" by (rule notnotD)
qed

lemma "A ∨ ¬A"
proof cases
  assume "A" thus ?thesis ..
next
  assume "¬A" thus ?thesis ..
qed
  

lemma "(¬(A ∧ B)) = (¬A ∨ ¬B)"
proof
  {
  assume 1: "¬(A ∧ B)"
  have 2: "¬A ∨ A" by (rule excluded_middle)
  thus "¬A ∨ ¬B"
    proof (rule disjE)
      { assume 3: "¬A"
        thus  "¬A ∨ ¬B" by (rule disjI1) }
    next
      { assume 4: "A"
        have 5: "¬B ∨ B" by (rule excluded_middle)
        thus "¬A ∨ ¬B"
        proof (rule disjE)
          {assume 6: "¬B"
            thus "¬A ∨ ¬B" by (rule disjI2) }
          next
          {assume 7: "B"
             from 4 and 7 have 8: "A ∧ B" by (rule conjI)
             from 1 and 8 show "¬A ∨ ¬B" by (rule notE) }
        qed
      }      
    qed
  }
next
  { assume 1: "¬A ∨ ¬B"
    show "¬(A ∧ B)"
    proof (rule notI)
      assume 2: "A ∧ B"
      from 1 have 3:"¬A ∨ ¬B" .
      moreover
      { assume 4: "¬A"
        from 2 have 5:"A" by (rule conjunct1)
        from 4 and 5 have False by (rule notE) }
      moreover
      { assume 6: "¬B"
        from 2 have 7: "B" by (rule conjunct2)
        from 6 and 7 have False by (rule notE) }
      ultimately show False by (rule disjE)        
    qed

  }
qed

end

RA_Relacion_2

theory Relacion_2 imports Main begin

(* Demostrar o refutar los siguientes lemas usando sólo las reglas básicas

  de deducción natural de la lógica proposicional y de la lógica de 
  predicados (allI, allE, exI y exE). *)

lemma "(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)" proof (rule impI)

 assume 1: "(∃x. ∀y. P x y)"
 show "(∀y. ∃x. P x y)"
 proof (rule allI)
   from 1 obtain x where 2: "∀y. P x y" by (rule exE)
   fix y
   from 2 have 3: "P x y" by (rule allE)
   from 3 show 4: "∃x. P x y" by (rule exI)
 qed

qed

lemma "(∀x. P x ⟶ Q) = ((∃x. P x) ⟶ Q)" proof

{ assume 1: "∀x. P x ⟶ Q"
  show "(∃x. P x) ⟶ Q"
  proof
    assume "∃x. P x"
    then obtain a where "P a" by (rule exE)
    have "P a ⟶ Q" using 1 by (rule allE)
    thus "Q" using `P a` by (rule mp)
  qed }

next

{ assume 2: "(∃x. P x) ⟶ Q"
  show "∀x. P x ⟶ Q"
  proof
    fix a
    show "P a ⟶ Q"
    proof
      assume "P a"
      hence 3: "∃x. P x" by (rule exI)
      from 2 and 3 show "Q" by (rule mp)
    qed
  qed }

qed

lemma "((∀ x. P x) ∧ (∀ x. Q x)) = (∀ x. (P x ∧ Q x))" proof {

 assume 1: "(∀ x. P x) ∧ (∀ x. Q x)"
 show "(∀ x. (P x ∧ Q x))"
   proof (rule allI)
     from 1 have 2: "(∀ x. P x)" by (rule conjE)
     from 1 have 3: "(∀ x. Q x)" by (rule conjE)
     fix a
     from 2 have 4: "P a" by (rule allE)
     from 3 have 5: "Q a" by (rule allE)
     from 4 and 5 show 6: "(P a) ∧ (Q a)" by (rule conjI)
   qed

} next {

 assume 7: "∀ x. (P x ∧ Q x)"
 show "(∀ x. P x) ∧ (∀ x. Q x)"
   proof
     fix a
     from 7 have 8: "P a ∧ Q a" by (rule allE)
     from 8 have 9: "P a" by (rule conjE)
     (* seguir... *)
     from 8 have 11: "Q a" by (rule allE)
     (* seguir... *)
   qed

} qed

oops

lemma "((∀ x. P x) ∨ (∀ x. Q x)) = (∀ x. (P x ∨ Q x))" oops

lemma "((∃ x. P x) ∨ (∃ x. Q x)) = (∃ x. (P x ∨ Q x))" oops

lemma "(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)" oops (* proof (rule impI)

 assume 1: "∀x. ∃y. P x y"
 show "∃y. ∀x. P x y"
   proof (rule exI)
     show " ∀x. P x y"
       proof (rule allI)
         fix x
         from 1 have 2: "∃y. P x y" by (rule allE)
         from 2 obtain y where 3: "P x y" by (rule exE)
         thus ?thesis ..
       qed

• )

lemma "(¬ (∀ x. P x)) = (∃ x. ¬ P x)" proof {

 assume 1: "¬ (∀ x. P x)" 
 show "∃ x. ¬ P x"
   proof
     

   qed

} next {

 assume 2: "∃ x. ¬ P x"
 show "¬ (∀ x. P x)"
   proof
     
   qed

} qed

end