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doc-src/TutorialI/Recdef/termination.thy

author | nipkow |

Wed Dec 13 09:39:53 2000 +0100 (2000-12-13) | |

changeset 10654 | 458068404143 |

parent 10522 | ed3964d1f1a4 |

child 10795 | 9e888d60d3e5 |

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

*** empty log message ***

1 (*<*)

2 theory termination = examples:

3 (*>*)

5 text{*

6 When a function is defined via \isacommand{recdef}, Isabelle tries to prove

7 its termination with the help of the user-supplied measure. All of the above

8 examples are simple enough that Isabelle can prove automatically that the

9 measure of the argument goes down in each recursive call. As a result,

10 $f$@{text".simps"} will contain the defining equations (or variants derived

11 from them) as theorems. For example, look (via \isacommand{thm}) at

12 @{thm[source]sep.simps} and @{thm[source]sep1.simps} to see that they define

13 the same function. What is more, those equations are automatically declared as

14 simplification rules.

16 In general, Isabelle may not be able to prove all termination conditions

17 (there is one for each recursive call) automatically. For example,

18 termination of the following artificial function

19 *}

21 consts f :: "nat\<times>nat \<Rightarrow> nat";

22 recdef f "measure(\<lambda>(x,y). x-y)"

23 "f(x,y) = (if x \<le> y then x else f(x,y+1))";

25 text{*\noindent

26 is not proved automatically (although maybe it should be). Isabelle prints a

27 kind of error message showing you what it was unable to prove. You will then

28 have to prove it as a separate lemma before you attempt the definition

29 of your function once more. In our case the required lemma is the obvious one:

30 *}

32 lemma termi_lem: "\<not> x \<le> y \<Longrightarrow> x - Suc y < x - y";

34 txt{*\noindent

35 It was not proved automatically because of the special nature of @{text"-"}

36 on @{typ"nat"}. This requires more arithmetic than is tried by default:

37 *}

39 apply(arith);

40 done

42 text{*\noindent

43 Because \isacommand{recdef}'s termination prover involves simplification,

44 we include with our second attempt the hint to use @{thm[source]termi_lem} as

45 a simplification rule:\indexbold{*recdef_simp}

46 *}

48 consts g :: "nat\<times>nat \<Rightarrow> nat";

49 recdef g "measure(\<lambda>(x,y). x-y)"

50 "g(x,y) = (if x \<le> y then x else g(x,y+1))"

51 (hints recdef_simp: termi_lem)

53 text{*\noindent

54 This time everything works fine. Now @{thm[source]g.simps} contains precisely

55 the stated recursion equation for @{term g} and they are simplification

56 rules. Thus we can automatically prove

57 *}

59 theorem "g(1,0) = g(1,1)";

60 apply(simp);

61 done

63 text{*\noindent

64 More exciting theorems require induction, which is discussed below.

66 If the termination proof requires a new lemma that is of general use, you can

67 turn it permanently into a simplification rule, in which case the above

68 \isacommand{hint} is not necessary. But our @{thm[source]termi_lem} is not

69 sufficiently general to warrant this distinction.

71 The attentive reader may wonder why we chose to call our function @{term g}

72 rather than @{term f} the second time around. The reason is that, despite

73 the failed termination proof, the definition of @{term f} did not

74 fail, and thus we could not define it a second time. However, all theorems

75 about @{term f}, for example @{thm[source]f.simps}, carry as a precondition

76 the unproved termination condition. Moreover, the theorems

77 @{thm[source]f.simps} are not simplification rules. However, this mechanism

78 allows a delayed proof of termination: instead of proving

79 @{thm[source]termi_lem} up front, we could prove

80 it later on and then use it to remove the preconditions from the theorems

81 about @{term f}. In most cases this is more cumbersome than proving things

82 up front.

83 %FIXME, with one exception: nested recursion.

84 *}

86 (*<*)

87 end

88 (*>*)