| author | haftmann | 
| Sun, 08 Oct 2017 22:28:20 +0200 | |
| changeset 66803 | dd8922885a68 | 
| parent 64242 | 93c6f0da5c70 | 
| child 67406 | 23307fd33906 | 
| permissions | -rw-r--r-- | 
| 16417 | 1 | theory Force imports Main begin | 
| 11456 | 2 | (*Use Divides rather than Main to experiment with the first proof. | 
| 3 | Otherwise, it is done by the nat_divide_cancel_factor simprocs.*) | |
| 11407 | 4 | |
| 58860 | 5 | declare div_mult_self_is_m [simp del] | 
| 10295 | 6 | |
| 11407 | 7 | lemma div_mult_self_is_m: | 
| 8 | "0<n \<Longrightarrow> (m*n) div n = (m::nat)" | |
| 64242 
93c6f0da5c70
more standardized theorem names for facts involving the div and mod identity
 haftmann parents: 
58860diff
changeset | 9 | apply (insert div_mult_mod_eq [of "m*n" n]) | 
| 11407 | 10 | apply simp | 
| 11 | done | |
| 10295 | 12 | |
| 13 | ||
| 14 | lemma "(\<forall>x. P x) \<and> (\<exists>x. Q x) \<longrightarrow> (\<forall>x. P x \<and> Q x)" | |
| 15 | apply clarify | |
| 16 | oops | |
| 17 | ||
| 18 | text {*
 | |
| 19 | proof\ {\isacharparenleft}prove{\isacharparenright}{\isacharcolon}\ step\ {\isadigit{1}}\isanewline
 | |
| 20 | \isanewline | |
| 21 | goal\ {\isacharparenleft}lemma{\isacharparenright}{\isacharcolon}\isanewline
 | |
| 22 | {\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
 | |
| 23 | \ {\isadigit{1}}{\isachardot}\ {\isasymAnd}x\ xa{\isachardot}\ {\isasymlbrakk}{\isasymforall}x{\isachardot}\ P\ x{\isacharsemicolon}\ Q\ xa{\isasymrbrakk}\ {\isasymLongrightarrow}\ P\ x\ {\isasymand}\ Q\ x
 | |
| 58860 | 24 | *} | 
| 10295 | 25 | |
| 26 | text {*
 | |
| 27 | couldn't find a good example of clarsimp | |
| 28 | ||
| 29 | @{thm[display]"someI"}
 | |
| 30 | \rulename{someI}
 | |
| 58860 | 31 | *} | 
| 10295 | 32 | |
| 33 | 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)" | |
| 34 | apply (rule someI) | |
| 35 | oops | |
| 36 | ||
| 37 | 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)" | |
| 38 | apply (fast intro!: someI) | |
| 39 | done | |
| 40 | ||
| 41 | text{*
 | |
| 42 | proof\ {\isacharparenleft}prove{\isacharparenright}{\isacharcolon}\ step\ \isadigit{1}\isanewline
 | |
| 43 | \isanewline | |
| 44 | goal\ {\isacharparenleft}lemma{\isacharparenright}{\isacharcolon}\isanewline
 | |
| 45 | {\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
 | |
| 46 | \ \isadigit{1}{\isachardot}\ {\isasymlbrakk}Q\ a{\isacharsemicolon}\ P\ a{\isasymrbrakk}\ {\isasymLongrightarrow}\ P\ {\isacharquery}x\ {\isasymand}\ Q\ {\isacharquery}x
 | |
| 47 | *} | |
| 48 | ||
| 49 | end | |
| 50 |