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

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 explicita)"
lemma I_d3: "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


-- "El ejercicio anterior hecho de un modo
automático (dado que no es posible hacerlo
por auto, es un buen candidato a que lo
resuelvan otros demostradores automáticos
proposicionales)"
lemma
  assumes 1:"A ∧ B" 
  shows  "B ∧ A" 
by (metis 1) 



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

-- "El anterior lemma
demostrado de forma aún más breve"
lemma "A ∧ B ⟶ A ∨ B"
by simp



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 version2: "(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


-- La versión anterior de la propiedad asociativa de la
disyunción (es decir, en forma de proposición implicativa)
demostrada de forma muy automática y breve.
lemma version2: "(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)"
by auto

-- Otra versión. la sentencia "note" no es necesaria.

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

-- "Éste también se puede
obtener de forma totalmente
automática"
lemma K: "A ⟶ (B ⟶ A)"
by auto



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


-- "Por simplificación es inmediata
su justificación"
lemma "(A ⟶ (B ⟶C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))"
by auto




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


-- "También se puede obtener directamente
por simplificación"
lemma "(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))"
by simp



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

-- "Está claro que éste sale
por auto y por simp"
lemma "¬¬A ⟶ A"
by simp



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"
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

-- "Éste sale por auto pero no
por el método simp"
lemma "((A ⟶ B) ⟶ A) ⟶ A"
by auto


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

-- "Demostración aún más corta
y automática"
lemma "A ∨ ¬A"
by simp


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

-- "Si no queremos entrar en detalles
lo podríamos mandar a la 'maza'"
lemma "(¬(A ∧ B)) = (¬A ∨ ¬B)"
by metis

end