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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["Ejercicios resueltos", "Subtitle"],

Cell[CellGroupData[{

Cell["Ejercicio 1. ", "Subsection",
  FontSize->16],

Cell[TextData[{
  "Dada las siguientes formas enunciativas:\n\tA: p \[UpArrow] (q \
\[LeftRightArrow] r)\n\tB: (p \[RightArrow] q) ",
  Cell[BoxData[
      \(TraditionalForm\`\[Wedge]\)]],
  " (r \[DownArrow] (~q))\nCalcular sus formas normales."
}], "Text",
  FontSize->14],

Cell[CellGroupData[{

Cell["Soluci\[OAcute]n:  ", "Subsubsection",
  CellDingbat->None,
  FontFamily->"Times New Roman",
  FontSize->16,
  FontWeight->"Bold",
  FontSlant->"Italic"],

Cell["\<\
Para calcular las formas normales de A primero introducimos la conectivas \
\[UpArrow] y \[LeftRightArrow] que hemos definido en el cap\[IAcute]culo \
anterior\
\>", "Text",
  FontSize->14],

Cell[CellGroupData[{

Cell[BoxData[{
    \(NAND[p_, q_] := \(! \((p && q)\)\)\), "\n", 
    \(Sii[a_, b_] := 
      Implies[a, b]\  && \ Implies[b, a]\), "\[IndentingNewLine]", 
    \(\(n = 3;\)\), "\n", 
    \(\(cadena = "\<\>";\)\), "\n", 
    \(\(cadena2 = "\<\>";\)\), "\n", 
    \(\(cad = "\<\>";\)\), "\n", 
    \(\(cad2 = "\<\>";\)\), "\n", 
    \(\(contradiccion = True;\)\), "\n", 
    \(\(tautologia = True;\)\), "\n", 
    \(\(p = Table[False, {t, n}];\)\), "\n", 
    \(\(expresion := 
        NAND[p[\([1]\)], Sii[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) \[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) \[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)"\),
      SequenceForm[ 
      "No es tautolog\[IAcute]a y la forma normal conmutativa es: ", 
        "((~p1) \[Or] (~p2) \[Or] (~p3)) \[And] ((~p1) \[Or] p2 \[Or] p3)"],
      Editable->False]], "Print"]
}, Closed]],

Cell[TextData[StyleBox["Para las formas normales de B s\[OAcute]lo hemos de \
definir la conectiva NOR[]:\t",
  FontFamily->"Times New Roman",
  FontSize->14]], "Text"],

Cell[CellGroupData[{

Cell[BoxData[{
    \(NOR[p_, q_] := \(! \((p || q)\)\)\), "\[IndentingNewLine]", 
    \(\(n = 3;\)\), "\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[\([2]\)]] && 
          NOR[p[\([3]\)], \(! 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 \[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))"],
      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) \[And] (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) \[And] (p1 \[Or] (~p2) \[Or] (~p3)) \[And] \
(p1 \[Or] p2 \[Or] (~p3)) \[And] (p1 \[Or] p2 \[Or] p3)"],
      Editable->False]], "Print"]
}, Closed]]
}, Closed]]
}, Open  ]],

Cell[CellGroupData[{

Cell["Ejercicio 2. ", "Subsection",
  FontSize->16],

Cell[TextData[{
  "Demostrar usando ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " que el conjunto {\[UpArrow]} y {\[DownArrow]} son conjuntos adecuados de \
conectivas."
}], "Text",
  FontSize->14],

Cell[CellGroupData[{

Cell["Soluci\[OAcute]n:  ", "Subsubsection",
  CellDingbat->None,
  FontFamily->"Times New Roman",
  FontSize->16,
  FontWeight->"Bold",
  FontSlant->"Italic"],

Cell[TextData[{
  "Para ver que el conjunto {\[UpArrow]} es un conjunto adecuado de \
conectivas hemos de probar que se tienen las siguientes equivalencias l\
\[OAcute]gicas:\n\ta) (\[Tilde]p) \[DoubleLeftRightArrow] (p \[UpArrow] p)\n\t\
b) (p ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\[Vee]\)\(\ \)\)\)]],
  "q) \[DoubleLeftRightArrow] (p \[UpArrow] p) \[UpArrow] (q \[UpArrow] q)\n\t\
c) (p ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\[Wedge]\)\(\ \)\)\)]],
  "q) \[DoubleLeftRightArrow] (p \[UpArrow] q) \[UpArrow] (p \[UpArrow] q)\n\
Veamos si lo son: "
}], "Text",
  FontSize->14],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(Sii[a_, b_] := Implies[a, b] && Implies[b, a];\)\), "\n", 
    \(\(NAND[p_, q_] := \(! \((p && q)\)\);\)\), "\[IndentingNewLine]", 
    \(\(n = 1;\)\), "\n", 
    \(\(tautologia = True;\)\), "\n", 
    \(\(p = Table[False, {t, n}];\)\), "\n", 
    \(\(expresion = 
        Sii[\(! p[\([1]\)]\), NAND[p[\([1]\)], 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, a) es una equivalencia l\[OAcute]gica.", "Text",
  FontSize->14],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(Sii[a_, b_] := Implies[a, b] && Implies[b, a];\)\), "\n", 
    \(\(NAND[p_, q_] := \(! \((p && q)\)\);\)\), "\[IndentingNewLine]", 
    \(\(n = 2;\)\), "\n", 
    \(\(tautologia = True;\)\), "\n", 
    \(\(p = Table[False, {t, n}];\)\), "\n", 
    \(\(expresion = 
        Sii[p[\([1]\)] || p[\([2]\)], 
          NAND[NAND[p[\([1]\)], p[\([1]\)]], 
            NAND[p[\([2]\)], p[\([2]\)]]]];\)\), "\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["\<\
De esta forma deducimos que b) tambi\[EAcute]n es una equivalencia \
l\[OAcute]gica.\
\>", "Text",
  FontSize->14],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(Sii[a_, b_] := Implies[a, b] && Implies[b, a];\)\), "\n", 
    \(\(NAND[p_, q_] := \(! \((p && q)\)\);\)\), "\[IndentingNewLine]", 
    \(\(n = 2;\)\), "\n", 
    \(\(tautologia = True;\)\), "\n", 
    \(\(p = Table[False, {t, n}];\)\), "\n", 
    \(\(expresion = 
        Sii[p[\([1]\)] && p[\([2]\)], 
          NAND[NAND[p[\([1]\)], p[\([2]\)]], 
            NAND[p[\([1]\)], p[\([2]\)]]]];\)\), "\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["\<\
As\[IAcute] c) tambi\[EAcute]n lo es y por tanto {\[UpArrow]} es un conjunto \
adecuado de conectivas.\
\>", "Text",
  FontSize->14],

Cell[TextData[{
  "De la misma forma podemos demostrar que el conjunto {\[DownArrow]} es un \
conjunto adecuado de conectivas. Para ello hemos de probar que:\n\ta) (\
\[Tilde]p) \[DoubleLeftRightArrow] (p \[DownArrow] p)\n\tb) (p ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\[Wedge]\)\(\ \)\)\)]],
  "q) \[DoubleLeftRightArrow] (p \[DownArrow] p) \[DownArrow] (q \[DownArrow] \
q)\n\tc) (p ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\[Vee]\)\(\ \)\)\)]],
  "q) \[DoubleLeftRightArrow] (p \[DownArrow] q) \[DownArrow] (p \[DownArrow] \
q)\nson equivalencias l\[OAcute]gicas. Veamoslo: "
}], "Text",
  FontSize->14],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(Sii[a_, b_] := Implies[a, b] && Implies[b, a];\)\), "\n", 
    \(\(NOR[p_, q_] := \(! \((p || q)\)\);\)\), "\[IndentingNewLine]", 
    \(\(n = 1;\)\), "\n", 
    \(\(tautologia = True;\)\), "\n", 
    \(\(p = Table[False, {t, n}];\)\), "\n", 
    \(\(expresion = 
        Sii[\(! p[\([1]\)]\), NOR[p[\([1]\)], 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, a) es una equivalencia l\[OAcute]gica.", "Text",
  FontSize->14],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(Sii[a_, b_] := Implies[a, b] && Implies[b, a];\)\), "\n", 
    \(\(NAND[p_, q_] := \(! \((p || q)\)\);\)\), "\[IndentingNewLine]", 
    \(\(n = 2;\)\), "\n", 
    \(\(tautologia = True;\)\), "\n", 
    \(\(p = Table[False, {t, n}];\)\), "\n", 
    \(\(expresion = 
        Sii[p[\([1]\)] || p[\([2]\)], 
          NOR[NOR[p[\([1]\)], p[\([2]\)]], 
            NOR[p[\([1]\)], p[\([2]\)]]]];\)\), "\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["\<\
De esta forma deducimos que b) tambi\[EAcute]n es una equivalencia \
l\[OAcute]gica.\
\>", "Text",
  FontSize->14],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(Sii[a_, b_] := Implies[a, b] && Implies[b, a];\)\), "\n", 
    \(\(NAND[p_, q_] := \(! \((p || q)\)\);\)\), "\[IndentingNewLine]", 
    \(\(n = 2;\)\), "\n", 
    \(\(tautologia = True;\)\), "\n", 
    \(\(p = Table[False, {t, n}];\)\), "\n", 
    \(\(expresion = 
        Sii[p[\([1]\)] && p[\([2]\)], 
          NOR[NOR[p[\([1]\)], p[\([1]\)]], 
            NOR[p[\([2]\)], p[\([2]\)]]]];\)\), "\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["\<\
As\[IAcute] c) tambi\[EAcute]n lo es y por tanto {\[DownArrow]} es un \
conjunto adecuado de conectivas.\
\>", "Text",
  FontSize->14]
}, Closed]]
}, Open  ]],

Cell[CellGroupData[{

Cell["Ejercicio 3. ", "Subsection",
  FontSize->16],

Cell["\<\
Un famoso adivino fue preguntado por el n\[UAcute]mero de alumnos que aprobar\
\[IAcute]a la asignatura de \[CapitalAAcute]lgebra y di\[OAcute] las \
siguientes pistas:
\t\[Bullet]  Si el alumno es hombre y naci\[OAcute] en un mes par entonces \
aprobar\[AAcute] el examen de teor\[IAcute]a.
\t\[Bullet]  Si el alumno es mujer y su DNI termina en 0, 3, 7, 8 o 9 \
entonces superar\[AAcute] las pr\[AAcute]cticas.
\t\[Bullet]  Si la edad del alumno es igual o superior a 20 a\[NTilde]os \
entonces superar\[AAcute] las pr\[AAcute]cticas y no aprobar\[AAcute] la teor\
\[IAcute]a.
\t\[Bullet]  El resto de alumnos aprobar\[AAcute] s\[OAcute]lo un de los dos \
ex\[AAcute]menes, el de pr\[AAcute]cticas o el de teor\[IAcute]a.
Sabiendo que para aprobar la asignatura de \[CapitalAAcute]lgebra hay que \
superar las pr\[AAcute]cticas y la teor\[IAcute]a, \[DownQuestion]aprobar\
\[AAcute] \[CapitalAAcute]lgebra un  alumno que naci\[OAcute] en junio de \
1975 y cuyo DNI termina en 9? \[DownQuestion]Y una alumna nacida en febrero \
de 1987 y con DNI 88256632?\
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\"El alumno tiene DNI terminado en 0, 3, 7, 8 o 9\"\n\tD = \"La edad del \
alumno es igual o superior a 20 a\[NTilde]os\"\n\tE = \"El alumno aprobar\
\[AAcute] el examen de teor\[IAcute]a\"\n\tF = \"El alumno aprobar\[AAcute] \
el examen de pr\[AAcute]cticas\"\n\tG = \"El alumno supera la asignatura de \
\[CapitalAAcute]lgebra\"\nCon estos nombres, lo que nos dijo el adivino se \
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