Re: [isabelle] Isar Proof



On 18/06/2008, at 8:11 AM, TIMOTHY KREMANN wrote:

> lemma
> "[| ALL x . (P (x::'a)  =>  ~ Q x)  /\
>                (P x        => ~ R x) /\
>               (Q x       =>  ~ R x);
> EX x .  P x;
> EX x .  Q x;
> EX x .  R x|] ==>
> EX  x y z. (~((x::'a) = y)  /\
>          ~( x      = z)  /\
>          ~( y      = z))"
>  apply(elim ex_forward)
>  apply(auto)
> done
>


Perhaps something like:

lemma
   assumes
     a1: "\<forall> x. (P (x::'a) \<longrightarrow> \<not> Q x) \<and>  
(P x  \<longrightarrow> \<not> R x) \<and> (Q x \<longrightarrow>  
\<not> R x)" and
     a2: "\<exists> x. P x" and
     a3: "\<exists> x. Q x" and
     a4: "\<exists> x. R x"
   shows
     "\<exists> x y z. (((x::'a) \<noteq> y) \<and> (x \<noteq> z)  
\<and> (y \<noteq> z))"
proof -
   from a2 obtain a where b1: "P a"
     by (auto)
   from a3 obtain b where b2: "Q b"
     by (auto)
   from a4 obtain c where b3: "R c"
     by (auto)
   from a1 b1 b2 have "a \<noteq> b"
     by (auto)
   moreover
   from a1 b1 b3 have "a \<noteq> c"
     by (auto)
   moreover
   from a1 b2 b3 have "b \<noteq> c"
     by (auto)
   ultimately
   show ?thesis
     by (auto)
qed



------------------------------------------------------------------
Dr Brendan Mahony
C3I Division                                    ph +61 8 8259 6046
Defence Science and Technology Organisation     fx +61 8 8259 5589
Edinburgh, South Australia      Brendan.Mahony at dsto.defence.gov.au

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