doc-src/TutorialI/Rules/Force.thy
author blanchet
Mon, 19 Apr 2010 18:14:45 +0200
changeset 36230 43d10a494c91
parent 16417 9bc16273c2d4
child 42637 381fdcab0f36
permissions -rw-r--r--
added warning about inconsistent context to Metis; it makes more sense here than in Sledgehammer, because Sledgehammer is unsound and there's no point in having people panicking about the consistency of their context when their context is in fact consistent

(* ID:         $Id$ *)
theory Force imports Main begin 
  (*Use Divides rather than Main to experiment with the first proof.
    Otherwise, it is done by the nat_divide_cancel_factor simprocs.*)

declare div_mult_self_is_m [simp del];

lemma div_mult_self_is_m: 
      "0<n \<Longrightarrow> (m*n) div n = (m::nat)"
apply (insert mod_div_equality [of "m*n" n])
apply simp
done


lemma "(\<forall>x. P x) \<and> (\<exists>x. Q x) \<longrightarrow> (\<forall>x. P x \<and> Q x)"
apply clarify
oops

text {*
proof\ {\isacharparenleft}prove{\isacharparenright}{\isacharcolon}\ step\ {\isadigit{1}}\isanewline
\isanewline
goal\ {\isacharparenleft}lemma{\isacharparenright}{\isacharcolon}\isanewline
{\isacharparenleft}{\isasymforall}x{\isachardot}\ P\ x{\isacharparenright}\ {\isasymand}\ {\isacharparenleft}{\isasymexists}x{\isachardot}\ Q\ x{\isacharparenright}\ {\isasymlongrightarrow}\ {\isacharparenleft}{\isasymforall}x{\isachardot}\ P\ x\ {\isasymand}\ Q\ x{\isacharparenright}\isanewline
\ {\isadigit{1}}{\isachardot}\ {\isasymAnd}x\ xa{\isachardot}\ {\isasymlbrakk}{\isasymforall}x{\isachardot}\ P\ x{\isacharsemicolon}\ Q\ xa{\isasymrbrakk}\ {\isasymLongrightarrow}\ P\ x\ {\isasymand}\ Q\ x
*};

text {*
couldn't find a good example of clarsimp

@{thm[display]"someI"}
\rulename{someI}
*};

lemma "\<lbrakk>Q a; P a\<rbrakk> \<Longrightarrow> P (SOME x. P x \<and> Q x) \<and> Q (SOME x. P x \<and> Q x)"
apply (rule someI)
oops

lemma "\<lbrakk>Q a; P a\<rbrakk> \<Longrightarrow> P (SOME x. P x \<and> Q x) \<and> Q (SOME x. P x \<and> Q x)"
apply (fast intro!: someI)
done

text{*
proof\ {\isacharparenleft}prove{\isacharparenright}{\isacharcolon}\ step\ \isadigit{1}\isanewline
\isanewline
goal\ {\isacharparenleft}lemma{\isacharparenright}{\isacharcolon}\isanewline
{\isasymlbrakk}Q\ a{\isacharsemicolon}\ P\ a{\isasymrbrakk}\ {\isasymLongrightarrow}\ P\ {\isacharparenleft}SOME\ x{\isachardot}\ P\ x\ {\isasymand}\ Q\ x{\isacharparenright}\ {\isasymand}\ Q\ {\isacharparenleft}SOME\ x{\isachardot}\ P\ x\ {\isasymand}\ Q\ x{\isacharparenright}\isanewline
\ \isadigit{1}{\isachardot}\ {\isasymlbrakk}Q\ a{\isacharsemicolon}\ P\ a{\isasymrbrakk}\ {\isasymLongrightarrow}\ P\ {\isacharquery}x\ {\isasymand}\ Q\ {\isacharquery}x
*}

end