doc-src/TutorialI/Recdef/termination.thy
author nipkow
Wed, 12 Dec 2001 09:04:20 +0100
changeset 12473 f41e477576b9
parent 12332 aea72a834c85
child 12489 c92e38c3cbaa
permissions -rw-r--r--
*** empty log message ***

(*<*)
theory termination = examples:
(*>*)

text{*
When a function~$f$ is defined via \isacommand{recdef}, Isabelle tries to prove
its termination with the help of the user-supplied measure.  Each of the examples
above is simple enough that Isabelle can automatically prove that the
argument's measure decreases in each recursive call. As a result,
$f$@{text".simps"} will contain the defining equations (or variants derived
from them) as theorems. For example, look (via \isacommand{thm}) at
@{thm[source]sep.simps} and @{thm[source]sep1.simps} to see that they define
the same function. What is more, those equations are automatically declared as
simplification rules.

Isabelle may fail to prove the termination condition for some
recursive call.  Let us try the following artificial function:*}

consts f :: "nat\<times>nat \<Rightarrow> nat"
recdef (*<*)(permissive)(*>*)f "measure(\<lambda>(x,y). x-y)"
  "f(x,y) = (if x \<le> y then x else f(x,y+1))"

text{*\noindent This definition fails, and Isabelle prints an error message
showing you what it was unable to prove. You will then have to prove it as a
separate lemma before you attempt the definition of your function once
more. In our case the required lemma is the obvious one: *}

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

txt{*\noindent
It was not proved automatically because of the awkward behaviour of subtraction
on type @{typ"nat"}. This requires more arithmetic than is tried by default:
*}

apply(arith)
done

text{*\noindent
Because \isacommand{recdef}'s termination prover involves simplification,
we include in our second attempt a hint: the \attrdx{recdef_simp} attribute 
says to use @{thm[source]termi_lem} as a simplification rule.  
*}

(*<*)global consts f :: "nat\<times>nat \<Rightarrow> nat" (*>*)
recdef f "measure(\<lambda>(x,y). x-y)"
  "f(x,y) = (if x \<le> y then x else f(x,y+1))"
(hints recdef_simp: termi_lem)
(*<*)local(*>*)
text{*\noindent
This time everything works fine. Now @{thm[source]f.simps} contains precisely
the stated recursion equation for @{text f}, which has been turned into a
simplification rule.  Thus we can automatically prove results such as this one:
*}

theorem "f(1,0) = f(1,1)"
apply(simp)
done

text{*\noindent
More exciting theorems require induction, which is discussed below.

If the termination proof requires a new lemma that is of general use, you can
turn it permanently into a simplification rule, in which case the above
\isacommand{hint} is not necessary. But our @{thm[source]termi_lem} is not
sufficiently general to warrant this distinction.
*}
(*<*)
end
(*>*)