(************** Content-type: application/mathematica **************
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Notebook[{

Cell[CellGroupData[{
Cell["\<\
6. L\[OAcute]gica proposicional: Tautolog\[IAcute]as, contradicciones, formas \
normales, equivalencias e implicaciones l\[OAcute]gicas y argumentaciones\
\>", "Title",
  FontSize->36],

Cell["Ejemplos con Mathematica", "Subtitle"],

Cell[CellGroupData[{

Cell["1. Tautolog\[IAcute]a y contradicci\[OAcute]n", "Section",
  FontSize->18],

Cell[CellGroupData[{

Cell["Ejemplo 6.1.", "Subsection",
  FontSize->16],

Cell[TextData[{
  "Comprobar si la forma enunciativa: A: (",
  Cell[BoxData[
      \(TraditionalForm\`p\_1\)]],
  Cell[BoxData[
      \(TraditionalForm\`\[Vee]\)]],
  Cell[BoxData[
      \(TraditionalForm\`p\_2\)]],
  ") ",
  Cell[BoxData[
      \(TraditionalForm\`\[Wedge]\)]],
  " (",
  Cell[BoxData[
      \(TraditionalForm\`p\_3\)]],
  Cell[BoxData[
      \(TraditionalForm\`\[Vee]\)]],
  Cell[BoxData[
      \(TraditionalForm\`p\_4\)]],
  ") es tautolog\[IAcute]a o contradicci\[OAcute]n."
}], "Text",
  FontSize->14],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(n = 4;\)\), "\n", 
    \(\(tautologia = True;\)\), "\n", 
    \(\(p = Table[False, {t, n}];\)\), "\n", 
    \(\(expresion := \((p[\([1]\)] || p[\([2]\)])\) && \((p[\([3]\)] || 
              p[\([4]\)])\);\)\), "\n", 
    \(\(For[i = 0, 
        i < 2^n, \(i++\), \[IndentingNewLine]j = i; \[IndentingNewLine]For[
          f = n, f > 0, \(f--\), \[IndentingNewLine]resto = 
            Mod[j, 2]; \[IndentingNewLine]j = 
            Floor[j/2]; \[IndentingNewLine]If[resto \[Equal] 0, 
            p[\([f]\)] = True, 
            p[\([f]\)] = False];\[IndentingNewLine]]; \[IndentingNewLine]If[
          TrueQ[expresion], Null, 
          tautologia = False];\[IndentingNewLine]];\)\), "\n", 
    \(If[tautologia, 
      Print["\<La forma enunciativa es una tautolog\[IAcute]a\>"], 
      Print["\<La forma enunciativa no es una tautolog\[IAcute]a\>"]]\)}], \
"Input"],

Cell[BoxData[
    \("La forma enunciativa no es una tautolog\[IAcute]a"\)], "Print"]
}, Closed]],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(n = 4;\)\), "\n", 
    \(\(contradiccion = True;\)\), "\n", 
    \(\(p = Table[False, {t, n}];\)\), "\n", 
    \(\(expresion := \((p[\([1]\)] || p[\([2]\)])\) && \((p[\([3]\)] || 
              p[\([4]\)])\);\)\), "\n", 
    \(\(For[i = 0, 
        i < 2^n, \(i++\), \[IndentingNewLine]j = i; \[IndentingNewLine]For[
          f = n, f > 0, \(f--\), \[IndentingNewLine]resto = 
            Mod[j, 2]; \[IndentingNewLine]j = 
            Floor[j/2]; \[IndentingNewLine]If[resto \[Equal] 0, 
            p[\([f]\)] = True, 
            p[\([f]\)] = False];\[IndentingNewLine]]; \[IndentingNewLine]If[
          TrueQ[expresion], contradiccion = False];\n];\)\), "\n", 
    \(If[contradiccion, 
      Print["\<La forma enunciativa es una contradicci\[OAcute]n\>"], 
      Print["\<La forma enunciativa no es una contradicci\[OAcute]n\>"]]\)}], \
"Input"],

Cell[BoxData[
    \("La forma enunciativa no es una contradicci\[OAcute]n"\)], "Print"]
}, Closed]]
}, Closed]]
}, Closed]],

Cell[CellGroupData[{

Cell["2. Formas normales", "Section",
  FontSize->18],

Cell[CellGroupData[{

Cell["Ejemplo 6.2.", "Subsection",
  FontSize->16],

Cell[TextData[{
  "Calcular las formas normales de la forma enunciativa A: (",
  Cell[BoxData[
      \(TraditionalForm\`p\_1\)]],
  Cell[BoxData[
      \(TraditionalForm\`\[Vee]\)]],
  Cell[BoxData[
      \(TraditionalForm\`p\_2\)]],
  ") \[RightArrow] ",
  Cell[BoxData[
      \(TraditionalForm\`p\_3\)]],
  "."
}], "Text",
  FontSize->14],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(n = 3;\)\), "\n", 
    \(\(cadena = "\<\>";\)\), "\n", 
    \(\(cadena2 = "\<\>";\)\), "\n", 
    \(\(cad = "\<\>";\)\), "\n", 
    \(\(cad2 = "\<\>";\)\), "\n", 
    \(\(contradiccion = True;\)\), "\n", 
    \(\(tautologia = True;\)\), "\n", 
    \(\(p = Table[False, {t, n}];\)\), "\n", 
    \(\(expresion := 
        Implies[\((p[\([1]\)] || p[\([2]\)])\), p[\([3]\)]];\)\), "\n", 
    \(\(For[i = 0, i < 2^n, \(i++\), 
        j = i; \[IndentingNewLine]cad = "\<\>"; \[IndentingNewLine]cad2 = "\<\
\>"; \[IndentingNewLine]For[f = n, f > 0, \(f--\), 
          resto = Mod[j, 2]; \[IndentingNewLine]j = 
            Floor[j/2]; \[IndentingNewLine]If[resto \[Equal] 0, 
            p[\([f]\)] = True; \[IndentingNewLine]If[f \[Equal] n, 
              cad = StringJoin["\<p\>", ToString[f], cad], 
              cad = StringJoin["\<p\>", ToString[f], "\< \[And] \>", 
                  cad]]; \(If[f \[Equal] n, 
                cad2 = StringJoin["\<(~p\>", ToString[f], "\<)\>", cad2], 
                cad2 = StringJoin["\<(~p\>", 
                    ToString[f], "\<)\>", "\< \[Or] \>", cad2]];\), 
            p[\([f]\)] = False; \[IndentingNewLine]If[f \[Equal] n, 
              cad = StringJoin["\<(~p\>", ToString[f], "\<)\>", cad], 
              cad = StringJoin["\<(~p\>", 
                  ToString[f], "\<)\>", "\< \[And] \>", 
                  cad]]; \[IndentingNewLine]If[f \[Equal] n, 
              cad2 = StringJoin["\<p\>", ToString[f], cad2], 
              cad2 = StringJoin["\<p\>", ToString[f], "\< \[Or] \>", 
                  cad2]];];]; \[IndentingNewLine]If[TrueQ[expresion], 
          If[cadena \[Equal] "\<\>", 
            cadena = StringJoin[cadena, "\<(\>", cad, "\<)\>"], 
            cadena = 
              StringJoin[cadena, "\< \[Or] (\>", 
                cad, "\<)\>"]]; \[IndentingNewLine]\(contradiccion = 
              False;\), 
          If[cadena2 \[Equal] "\<\>", 
            cadena2 = StringJoin[cadena2, "\<(\>", cad2, "\<)\>"], 
            cadena2 = 
              StringJoin[cadena2, "\< \[And] (\>", 
                cad2, "\<)\>"]]; \[IndentingNewLine]tautologia = 
            False;];];\)\), "\n", 
    \(\(If[contradiccion, Print["\<Es una contradicci\[OAcute]n.\>"], 
        Print["\<No es contradicci\[OAcute]n y la forma normal disyuntiva es: \
\>", cadena]];\)\), "\n", 
    \(\(If[tautologia, Print["\<Es una tautolog\[IAcute]a.\>"], 
        Print["\<No es tautolog\[IAcute]a y la forma normal conmutativa es: \
\>", cadena2]];\)\)}], "Input"],

Cell[BoxData[
    InterpretationBox[\("No es contradicci\[OAcute]n y la forma normal \
disyuntiva es: "\[InvisibleSpace]"(p1 \[And] p2 \[And] p3) \[Or] (p1 \[And] \
(~p2) \[And] p3) \[Or] ((~p1) \[And] p2 \[And] p3) \[Or] ((~p1) \[And] (~p2) \
\[And] p3) \[Or] ((~p1) \[And] (~p2) \[And] (~p3))"\),
      SequenceForm[ 
      "No es contradicci\[OAcute]n y la forma normal disyuntiva es: ", 
        "(p1 \[And] p2 \[And] p3) \[Or] (p1 \[And] (~p2) \[And] p3) \[Or] \
((~p1) \[And] p2 \[And] p3) \[Or] ((~p1) \[And] (~p2) \[And] p3) \[Or] ((~p1) \
\[And] (~p2) \[And] (~p3))"],
      Editable->False]], "Print"],

Cell[BoxData[
    InterpretationBox[\("No es tautolog\[IAcute]a y la forma normal \
conmutativa es: "\[InvisibleSpace]"((~p1) \[Or] (~p2) \[Or] p3) \[And] ((~p1) \
\[Or] p2 \[Or] p3) \[And] (p1 \[Or] (~p2) \[Or] p3)"\),
      SequenceForm[ 
      "No es tautolog\[IAcute]a y la forma normal conmutativa es: ", 
        "((~p1) \[Or] (~p2) \[Or] p3) \[And] ((~p1) \[Or] p2 \[Or] p3) \[And] \
(p1 \[Or] (~p2) \[Or] p3)"],
      Editable->False]], "Print"]
}, Closed]]
}, Closed]],

Cell[CellGroupData[{

Cell["Ejemplo 6.3.", "Subsection",
  FontSize->16],

Cell[TextData[{
  "Calcular una forma enunciativa restringida equivalente a la forma \
enunciativa A: ",
  Cell[BoxData[
      \(TraditionalForm\`p\_1\)]],
  "\[RightArrow]",
  Cell[BoxData[
      \(TraditionalForm\`p\_2\)]],
  "."
}], "Text",
  FontSize->14],

Cell["\<\
Bastar\[AAcute] con dar la forma normal disyuntiva o la forma normal \
conjuntiva de A para tener tal forma enunciativa restringida.\
\>", "Text",
  FontSize->14],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(n = 2;\)\), "\n", 
    \(\(cadena = "\<\>";\)\), "\n", 
    \(\(cadena2 = "\<\>";\)\), "\n", 
    \(\(cad = "\<\>";\)\), "\n", 
    \(\(cad2 = "\<\>";\)\), "\n", 
    \(\(contradiccion = True;\)\), "\n", 
    \(\(tautologia = True;\)\), "\n", 
    \(\(p = Table[False, {t, n}];\)\), "\n", 
    \(\(expresion := Implies[p[\([1]\)], p[\([2]\)]];\)\), "\n", 
    \(\(For[i = 0, i < 2^n, \(i++\), 
        j = i; \[IndentingNewLine]cad = "\<\>"; \[IndentingNewLine]cad2 = "\<\
\>"; \[IndentingNewLine]For[f = n, f > 0, \(f--\), 
          resto = Mod[j, 2]; \[IndentingNewLine]j = 
            Floor[j/2]; \[IndentingNewLine]If[resto \[Equal] 0, 
            p[\([f]\)] = True; \[IndentingNewLine]If[f \[Equal] n, 
              cad = StringJoin["\<p\>", ToString[f], cad], 
              cad = StringJoin["\<p\>", ToString[f], "\< \[And] \>", 
                  cad]]; \(If[f \[Equal] n, 
                cad2 = StringJoin["\<(~p\>", ToString[f], "\<)\>", cad2], 
                cad2 = StringJoin["\<(~p\>", 
                    ToString[f], "\<)\>", "\< \[Or] \>", cad2]];\), 
            p[\([f]\)] = False; \[IndentingNewLine]If[f \[Equal] n, 
              cad = StringJoin["\<(~p\>", ToString[f], "\<)\>", cad], 
              cad = StringJoin["\<(~p\>", 
                  ToString[f], "\<)\>", "\< \[And] \>", 
                  cad]]; \[IndentingNewLine]If[f \[Equal] n, 
              cad2 = StringJoin["\<p\>", ToString[f], cad2], 
              cad2 = StringJoin["\<p\>", ToString[f], "\< \[Or] \>", 
                  cad2]];];]; \[IndentingNewLine]If[TrueQ[expresion], 
          If[cadena \[Equal] "\<\>", 
            cadena = StringJoin[cadena, "\<(\>", cad, "\<)\>"], 
            cadena = 
              StringJoin[cadena, "\< \[Or] (\>", 
                cad, "\<)\>"]]; \[IndentingNewLine]\(contradiccion = 
              False;\), 
          If[cadena2 \[Equal] "\<\>", 
            cadena2 = StringJoin[cadena2, "\<(\>", cad2, "\<)\>"], 
            cadena2 = 
              StringJoin[cadena2, "\< \[And] (\>", 
                cad2, "\<)\>"]]; \[IndentingNewLine]tautologia = 
            False;];];\)\), "\n", 
    \(\(If[contradiccion, Print["\<Es una contradicci\[OAcute]n.\>"], 
        Print["\<No es contradicci\[OAcute]n y la forma normal disyuntiva es: \
\>", cadena]];\)\), "\n", 
    \(\(If[tautologia, Print["\<Es una tautolog\[IAcute]a.\>"], 
        Print["\<No es tautolog\[IAcute]a y la forma normal conmutativa es: \
\>", cadena2]];\)\)}], "Input"],

Cell[BoxData[
    InterpretationBox[\("No es contradicci\[OAcute]n y la forma normal \
disyuntiva es: "\[InvisibleSpace]"(p1 \[And] p2) \[Or] ((~p1) \[And] p2) \
\[Or] ((~p1) \[And] (~p2))"\),
      SequenceForm[ 
      "No es contradicci\[OAcute]n y la forma normal disyuntiva es: ", 
        "(p1 \[And] p2) \[Or] ((~p1) \[And] p2) \[Or] ((~p1) \[And] (~p2))"],
      Editable->False]], "Print"],

Cell[BoxData[
    InterpretationBox[\("No es tautolog\[IAcute]a y la forma normal \
conmutativa es: "\[InvisibleSpace]"((~p1) \[Or] p2)"\),
      SequenceForm[ 
      "No es tautolog\[IAcute]a y la forma normal conmutativa es: ", 
        "((~p1) \[Or] p2)"],
      Editable->False]], "Print"]
}, Closed]],

Cell[TextData[{
  "Por tanto, una soluci\[OAcute]n para este problema ser\[IAcute]a la forma \
enunciativa restringida (",
  Cell[BoxData[
      \(TraditionalForm\`p\_1\)]],
  Cell[BoxData[
      \(TraditionalForm\`\[Vee]\)]],
  " (~",
  Cell[BoxData[
      \(TraditionalForm\`p\_2\)]],
  "))."
}], "Text",
  FontSize->14]
}, Closed]],

Cell[CellGroupData[{

Cell["Ejemplo 6.4.", "Subsection",
  FontSize->16],

Cell["\<\
Aplicamos el procedimiento logica a la forma enunciativa ((p\[RightArrow]q)\
\[RightArrow](~r))\
\>", "Text",
  FontSize->14],

Cell[BoxData[
    \(logica := 
      Module[{tabla, cadena, cadena2, cad, cad2, contradiccion, tautologia}, 
        tabla = Table["\<F\>", {x, \((2^n + 1)\)}, {y, 
              n + 1}]; \[IndentingNewLine]For[k = 1, k < n + 1, \(k++\), 
          tabla[\([1, 
                k]\)] = \\!\\("\<p\>" \\_ 
                  k\\)]; \[IndentingNewLine]tabla[\([1, 
              n + 1]\)] = "\<Exp\>"; \[IndentingNewLine]For[i = 0, 
          i < 2^n, \(i++\), 
          j = i; \[IndentingNewLine]For[f = n, f > 0, \(f--\), 
            resto = Mod[j, 2]; \[IndentingNewLine]j = 
              Floor[j/2]; \[IndentingNewLine]If[resto \[Equal] 0, 
              p[\([f]\)] = True; tabla[\([i + 2, f]\)] = "\<V\>", 
              p[\([f]\)] = False];]; \[IndentingNewLine]If[TrueQ[expresion], 
            tabla[\([i + 2, 
                  n + 1]\)] = "\<V\>"];]; \[IndentingNewLine]Print["\<La \
tabla de verdad es: \>"]; \[IndentingNewLine]Print[
          TableForm[
            tabla]]; \[IndentingNewLine]cadena = "\<\>"; \
\[IndentingNewLine]cadena2 = "\<\>"; \[IndentingNewLine]cad = "\<\>"; \
\[IndentingNewLine]cad2 = "\<\>"; \[IndentingNewLine]contradiccion = 
          True; \[IndentingNewLine]tautologia = True; \[IndentingNewLine]For[
          i = 0, i < 2^n, \(i++\), 
          j = i; \[IndentingNewLine]cad = "\<\>"; \[IndentingNewLine]cad2 = "\
\<\>"; \[IndentingNewLine]For[f = n, f > 0, \(f--\), 
            resto = Mod[j, 2]; \[IndentingNewLine]j = 
              Floor[j/2]; \[IndentingNewLine]If[resto \[Equal] 0, 
              p[\([f]\)] = True; \[IndentingNewLine]If[f \[Equal] n, 
                cad = StringJoin["\<p\>", ToString[f], cad], 
                cad = StringJoin["\<p\>", ToString[f], "\< \[And] \>", 
                    cad]]; \[IndentingNewLine]If[f \[Equal] n, 
                cad2 = StringJoin["\<(~p\>", ToString[f], "\<)\>", cad2], 
                cad2 = StringJoin["\<(~p\>", 
                    ToString[f], "\<)\>", "\< \[Or] \>", cad2]];, 
              p[\([f]\)] = False; \[IndentingNewLine]If[f \[Equal] n, 
                cad = StringJoin["\<(~p\>", ToString[f], "\<)\>", cad], 
                cad = StringJoin["\<(~p\>", 
                    ToString[f], "\<)\>", "\< \[And] \>", 
                    cad]]; \[IndentingNewLine]If[f \[Equal] n, 
                cad2 = StringJoin["\<p\>", ToString[f], cad2], 
                cad2 = StringJoin["\<p\>", ToString[f], "\< \[Or] \>", 
                    cad2]];];]; \[IndentingNewLine]If[TrueQ[expresion], 
            If[cadena \[Equal] "\<\>", 
              cadena = StringJoin[cadena, "\<(\>", cad, "\<)\>"], 
              cadena = 
                StringJoin[cadena, "\< \[Or] (\>", 
                  cad, "\<)\>"]]; \[IndentingNewLine]contradiccion = False;, 
            If[cadena2 \[Equal] "\<\>", 
              cadena2 = StringJoin[cadena2, "\<(\>", cad2, "\<)\>"], 
              cadena2 = 
                StringJoin[cadena2, "\< \[And] (\>", 
                  cad2, "\<)\>"]]; \[IndentingNewLine]tautologia = 
              False;];]; \[IndentingNewLine]If[contradiccion, 
          Print["\<Es una contradicci\[OAcute]n.\>"], 
          Print["\<No es contradicci\[OAcute]n y la forma normal disyuntiva \
es: \>", cadena]]; \[IndentingNewLine]If[tautologia, 
          Print["\<Es una tautolog\[IAcute]a.\>"], 
          Print["\<No es tautolog\[IAcute]a y la forma normal conjuntiva es: \
\>", cadena2]]]\)], "Input"],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(n = 3;\)\), "\n", 
    \(\(p = Table[False, {t, n}];\)\), "\n", 
    \(expresion := 
      Implies[Implies[p[\([1]\)], p[\([2]\)]], \(! p[\([3]\)]\)]\), "\n", 
    \(logica\)}], "Input"],

Cell[BoxData[
    \("La tabla de verdad es: "\)], "Print"],

Cell[BoxData[
    TagBox[GridBox[{
          {\("p"\_1\), \("p"\_2\), \("p"\_3\), "\<\"Exp\"\>"},
          {"\<\"V\"\>", "\<\"V\"\>", "\<\"V\"\>", "\<\"F\"\>"},
          {"\<\"V\"\>", "\<\"V\"\>", "\<\"F\"\>", "\<\"V\"\>"},
          {"\<\"V\"\>", "\<\"F\"\>", "\<\"V\"\>", "\<\"V\"\>"},
          {"\<\"V\"\>", "\<\"F\"\>", "\<\"F\"\>", "\<\"V\"\>"},
          {"\<\"F\"\>", "\<\"V\"\>", "\<\"V\"\>", "\<\"F\"\>"},
          {"\<\"F\"\>", "\<\"V\"\>", "\<\"F\"\>", "\<\"V\"\>"},
          {"\<\"F\"\>", "\<\"F\"\>", "\<\"V\"\>", "\<\"F\"\>"},
          {"\<\"F\"\>", "\<\"F\"\>", "\<\"F\"\>", "\<\"V\"\>"}
          },
        RowSpacings->1,
        ColumnSpacings->3,
        RowAlignments->Baseline,
        ColumnAlignments->{Left}],
      Function[ BoxForm`e$, 
        TableForm[ BoxForm`e$]]]], "Print"],

Cell[BoxData[
    InterpretationBox[\("No es contradicci\[OAcute]n y la forma normal \
disyuntiva es: "\[InvisibleSpace]"(p1 \[And] p2 \[And] (~p3)) \[Or] (p1 \
\[And] (~p2) \[And] p3) \[Or] (p1 \[And] (~p2) \[And] (~p3)) \[Or] ((~p1) \
\[And] p2 \[And] (~p3)) \[Or] ((~p1) \[And] (~p2) \[And] (~p3))"\),
      SequenceForm[ 
      "No es contradicci\[OAcute]n y la forma normal disyuntiva es: ", 
        "(p1 \[And] p2 \[And] (~p3)) \[Or] (p1 \[And] (~p2) \[And] p3) \[Or] \
(p1 \[And] (~p2) \[And] (~p3)) \[Or] ((~p1) \[And] p2 \[And] (~p3)) \[Or] \
((~p1) \[And] (~p2) \[And] (~p3))"],
      Editable->False]], "Print"],

Cell[BoxData[
    InterpretationBox[\("No es tautolog\[IAcute]a y la forma normal \
conjuntiva es: "\[InvisibleSpace]"((~p1) \[Or] (~p2) \[Or] (~p3)) \[And] (p1 \
\[Or] (~p2) \[Or] (~p3)) \[And] (p1 \[Or] p2 \[Or] (~p3))"\),
      SequenceForm[ 
      "No es tautolog\[IAcute]a y la forma normal conjuntiva es: ", 
        "((~p1) \[Or] (~p2) \[Or] (~p3)) \[And] (p1 \[Or] (~p2) \[Or] (~p3)) \
\[And] (p1 \[Or] p2 \[Or] (~p3))"],
      Editable->False]], "Print"]
}, Closed]]
}, Closed]]
}, Closed]],

Cell[CellGroupData[{

Cell["3. Equivalencias l\[OAcute]gicas e implicaciones l\[OAcute]gicas", \
"Section",
  FontSize->18],

Cell[CellGroupData[{

Cell["Ejemplo 6.5.", "Subsection",
  FontSize->16],

Cell["\<\
Comprobar que las siguientes formas enunciativas son equivalentes:
\ta) (p \[RightArrow] q)
\tb) ((~q) \[RightArrow] (~p))\
\>", "Text",
  FontSize->14],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(Sii[a_, b_] := Implies[a, b] && Implies[b, a];\)\), "\n", 
    \(\(n = 2;\)\), "\n", 
    \(\(tautologia = True;\)\), "\n", 
    \(\(p = Table[False, {t, n}];\)\), "\n", 
    \(\(expresion := 
        Sii[Implies[p[\([1]\)], p[\([2]\)]], 
          Implies[\(! p[\([2]\)]\), \(! p[\([1]\)]\)]];\)\), "\n", 
    \(\(For[i = 0, i < 2^n, \(i++\), 
        j = i; \[IndentingNewLine]For[f = n, f > 0, \(f--\), 
          resto = Mod[j, 2]; \[IndentingNewLine]j = 
            Floor[j/2]; \[IndentingNewLine]If[resto \[Equal] 0, 
            p[\([f]\)] = True, p[\([f]\)] = False];]; \[IndentingNewLine]If[
          TrueQ[expresion], Null, tautologia = False];];\)\), "\n", 
    \(If[tautologia, 
      Print["\<La forma enunciativa es una tautolog\[IAcute]a\>"], 
      Print["\<La forma enunciativa no es una tautolog\[IAcute]a\>"]]\)}], \
"Input"],

Cell[BoxData[
    \("La forma enunciativa es una tautolog\[IAcute]a"\)], "Print"]
}, Closed]],

Cell["\<\
Por tanto, la forma enunciativa ((p \[RightArrow] q) \[LeftRightArrow] ((~ q) \
\[RightArrow] (~ p))) es una tautolog\[IAcute]a de lo que deducimos que (p \
\[RightArrow] q) es l\[OAcute]gicamente equivalente a ((~ q) \[RightArrow] (~ \
p)).\
\>", "Text",
  FontSize->14]
}, Closed]]
}, Closed]],

Cell[CellGroupData[{

Cell["4. Argumentaciones v\[AAcute]lidas", "Section",
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S\[OAcute]lo tenemos que ver si la forma enunciativa (((p\[Wedge]q)\
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  "Cierta empresa necesita incrementar su plantilla con un nuevo miembro que \
pueda programar en ",
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\[OAcute]n entonces puede programar en ",
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una persona que sepa utilizar el programa ",
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otro caso ser\[AAcute] negativa."
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