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doc-src/TutorialI/Inductive/Star.thy

author | nipkow |

Wed Dec 06 13:22:58 2000 +0100 (2000-12-06) | |

changeset 10608 | 620647438780 |

parent 10520 | bb9dfcc87951 |

child 10898 | b086f4e1722f |

permissions | -rw-r--r-- |

*** empty log message ***

1 (*<*)theory Star = Main:(*>*)

3 section{*The reflexive transitive closure*}

5 text{*\label{sec:rtc}

6 Many inductive definitions define proper relations rather than merely set

7 like @{term even}. A perfect example is the

8 reflexive transitive closure of a relation. This concept was already

9 introduced in \S\ref{sec:Relations}, where the operator @{text"^*"} was

10 defined as a least fixed point because inductive definitions were not yet

11 available. But now they are:

12 *}

14 consts rtc :: "('a \<times> 'a)set \<Rightarrow> ('a \<times> 'a)set" ("_*" [1000] 999)

15 inductive "r*"

16 intros

17 rtc_refl[iff]: "(x,x) \<in> r*"

18 rtc_step: "\<lbrakk> (x,y) \<in> r; (y,z) \<in> r* \<rbrakk> \<Longrightarrow> (x,z) \<in> r*"

20 text{*\noindent

21 The function @{term rtc} is annotated with concrete syntax: instead of

22 @{text"rtc r"} we can read and write @{term"r*"}. The actual definition

23 consists of two rules. Reflexivity is obvious and is immediately given the

24 @{text iff} attribute to increase automation. The

25 second rule, @{thm[source]rtc_step}, says that we can always add one more

26 @{term r}-step to the left. Although we could make @{thm[source]rtc_step} an

27 introduction rule, this is dangerous: the recursion in the second premise

28 slows down and may even kill the automatic tactics.

30 The above definition of the concept of reflexive transitive closure may

31 be sufficiently intuitive but it is certainly not the only possible one:

32 for a start, it does not even mention transitivity explicitly.

33 The rest of this section is devoted to proving that it is equivalent to

34 the ``standard'' definition. We start with a simple lemma:

35 *}

37 lemma [intro]: "(x,y) : r \<Longrightarrow> (x,y) \<in> r*"

38 by(blast intro: rtc_step);

40 text{*\noindent

41 Although the lemma itself is an unremarkable consequence of the basic rules,

42 it has the advantage that it can be declared an introduction rule without the

43 danger of killing the automatic tactics because @{term"r*"} occurs only in

44 the conclusion and not in the premise. Thus some proofs that would otherwise

45 need @{thm[source]rtc_step} can now be found automatically. The proof also

46 shows that @{text blast} is quite able to handle @{thm[source]rtc_step}. But

47 some of the other automatic tactics are more sensitive, and even @{text

48 blast} can be lead astray in the presence of large numbers of rules.

50 To prove transitivity, we need rule induction, i.e.\ theorem

51 @{thm[source]rtc.induct}:

52 @{thm[display]rtc.induct}

53 It says that @{text"?P"} holds for an arbitrary pair @{text"(?xb,?xa) \<in>

54 ?r*"} if @{text"?P"} is preserved by all rules of the inductive definition,

55 i.e.\ if @{text"?P"} holds for the conclusion provided it holds for the

56 premises. In general, rule induction for an $n$-ary inductive relation $R$

57 expects a premise of the form $(x@1,\dots,x@n) \in R$.

59 Now we turn to the inductive proof of transitivity:

60 *}

62 lemma rtc_trans: "\<lbrakk> (x,y) \<in> r*; (y,z) \<in> r* \<rbrakk> \<Longrightarrow> (x,z) \<in> r*"

63 apply(erule rtc.induct)

65 txt{*\noindent

66 Unfortunately, even the resulting base case is a problem

67 @{subgoals[display,indent=0,goals_limit=1]}

68 and maybe not what you had expected. We have to abandon this proof attempt.

69 To understand what is going on, let us look again at @{thm[source]rtc.induct}.

70 In the above application of @{text erule}, the first premise of

71 @{thm[source]rtc.induct} is unified with the first suitable assumption, which

72 is @{term"(x,y) \<in> r*"} rather than @{term"(y,z) \<in> r*"}. Although that

73 is what we want, it is merely due to the order in which the assumptions occur

74 in the subgoal, which it is not good practice to rely on. As a result,

75 @{text"?xb"} becomes @{term x}, @{text"?xa"} becomes

76 @{term y} and @{text"?P"} becomes @{term"%u v. (u,z) : r*"}, thus

77 yielding the above subgoal. So what went wrong?

79 When looking at the instantiation of @{text"?P"} we see that it does not

80 depend on its second parameter at all. The reason is that in our original

81 goal, of the pair @{term"(x,y)"} only @{term x} appears also in the

82 conclusion, but not @{term y}. Thus our induction statement is too

83 weak. Fortunately, it can easily be strengthened:

84 transfer the additional premise @{prop"(y,z):r*"} into the conclusion:*}

85 (*<*)oops(*>*)

86 lemma rtc_trans[rule_format]:

87 "(x,y) \<in> r* \<Longrightarrow> (y,z) \<in> r* \<longrightarrow> (x,z) \<in> r*"

89 txt{*\noindent

90 This is not an obscure trick but a generally applicable heuristic:

91 \begin{quote}\em

92 Whe proving a statement by rule induction on $(x@1,\dots,x@n) \in R$,

93 pull all other premises containing any of the $x@i$ into the conclusion

94 using $\longrightarrow$.

95 \end{quote}

96 A similar heuristic for other kinds of inductions is formulated in

97 \S\ref{sec:ind-var-in-prems}. The @{text rule_format} directive turns

98 @{text"\<longrightarrow>"} back into @{text"\<Longrightarrow>"}. Thus in the end we obtain the original

99 statement of our lemma.

100 *}

102 apply(erule rtc.induct)

104 txt{*\noindent

105 Now induction produces two subgoals which are both proved automatically:

106 @{subgoals[display,indent=0]}

107 *}

109 apply(blast);

110 apply(blast intro: rtc_step);

111 done

113 text{*

114 Let us now prove that @{term"r*"} is really the reflexive transitive closure

115 of @{term r}, i.e.\ the least reflexive and transitive

116 relation containing @{term r}. The latter is easily formalized

117 *}

119 consts rtc2 :: "('a \<times> 'a)set \<Rightarrow> ('a \<times> 'a)set"

120 inductive "rtc2 r"

121 intros

122 "(x,y) \<in> r \<Longrightarrow> (x,y) \<in> rtc2 r"

123 "(x,x) \<in> rtc2 r"

124 "\<lbrakk> (x,y) \<in> rtc2 r; (y,z) \<in> rtc2 r \<rbrakk> \<Longrightarrow> (x,z) \<in> rtc2 r"

126 text{*\noindent

127 and the equivalence of the two definitions is easily shown by the obvious rule

128 inductions:

129 *}

131 lemma "(x,y) \<in> rtc2 r \<Longrightarrow> (x,y) \<in> r*"

132 apply(erule rtc2.induct);

133 apply(blast);

134 apply(blast);

135 apply(blast intro: rtc_trans);

136 done

138 lemma "(x,y) \<in> r* \<Longrightarrow> (x,y) \<in> rtc2 r"

139 apply(erule rtc.induct);

140 apply(blast intro: rtc2.intros);

141 apply(blast intro: rtc2.intros);

142 done

144 text{*

145 So why did we start with the first definition? Because it is simpler. It

146 contains only two rules, and the single step rule is simpler than

147 transitivity. As a consequence, @{thm[source]rtc.induct} is simpler than

148 @{thm[source]rtc2.induct}. Since inductive proofs are hard enough, we should

149 certainly pick the simplest induction schema available for any concept.

150 Hence @{term rtc} is the definition of choice.

152 \begin{exercise}\label{ex:converse-rtc-step}

153 Show that the converse of @{thm[source]rtc_step} also holds:

154 @{prop[display]"[| (x,y) : r*; (y,z) : r |] ==> (x,z) : r*"}

155 \end{exercise}

156 \begin{exercise}

157 Repeat the development of this section, but starting with a definition of

158 @{term rtc} where @{thm[source]rtc_step} is replaced by its converse as shown

159 in exercise~\ref{ex:converse-rtc-step}.

160 \end{exercise}

161 *}

162 (*<*)

163 lemma rtc_step2[rule_format]: "(x,y) : r* \<Longrightarrow> (y,z) : r --> (x,z) : r*"

164 apply(erule rtc.induct);

165 apply blast;

166 apply(blast intro:rtc_step)

167 done

169 end

170 (*>*)