[isabelle] nonnegative quadratic polynomial

[Omar Jasim]

Dear all,


Please I need to prove that for any nonnegative quadratic polynomial
equation the discriminant will be nonpositive. I found a proposition in a
text book bellow :
https://books.google.co.uk/books?id=TPE0fXGnYtMC&pg=PA81&lpg=PA81&dq=if+f(x)+%3E0+then+b%5E2-4ac%3C0&source=bl&ots=eUAdjNvJpT&sig=KwJxcX_5RqIsCL4wJ3nrfvfZe9g&hl=en&sa=X&ved=0ahUKEwjlxpjm2ZjNAhWMLsAKHX2vAs8Q6AEILzAE#v=onepage&q&f=false

but for this one, the equality hold iff x=-b/2*a and I don't want to put
this as an assumption as I need this lemma in another work. the lemma I'm
trying is:

lemma
fixes a b c x :: real
assumes "a > 0"
and "⋀x. a*x^2 + b*x + c≥0 "
shows " discrim a b c ≤0"

I proved that for strictly positive equation as bellow:

lemma
fixes a b c x :: real
assumes "a > 0"
and "⋀x. (a*(x)^2 + b*x + c) >0 "
shows " discrim a b c ≤0"
using assms by (metis discriminant_pos_ex less_le not_less)

but for nonnegative equation (the first lemma) I couldn't and way. please
could any one help me in this because I spend alot of time trying to prove
it but unfortunately I failed.

Omar

[Wenda Li]

Dear Omar,

Here you are,

lemma
fixes a b c x :: real
assumes "a > 0"
and "⋀x. a*x^2 + b*x + c≥0 "
shows "discrim a b c ≤0"
proof -
have "∃x. a*x^2 + b*x + c<0 " when "discrim a b c > 0" "a>0"
proof -
let ?P="λx. a*x^2 + b*x +c"
have "?P (- b / (2*a)) = (4*a*c-b^2)/(4*a)" using ‹a>0›
apply (auto simp add:field_simps)
by algebra
also have "... <0"
using that unfolding discrim_def by (simp add: divide_neg_pos)
finally show ?thesis by blast
qed
then show ?thesis using assms
using not_less by blast
qed

Best,
Wenda

Post by Omar Jasim
Dear all,
Please I need to prove that for any nonnegative quadratic polynomial
equation the discriminant will be nonpositive. I found a proposition in a
https://books.google.co.uk/books?id=TPE0fXGnYtMC&pg=PA81&lpg=PA81&dq=if+f(x)+%3E0+then+b%5E2-4ac%3C0&source=bl&ots=eUAdjNvJpT&sig=KwJxcX_5RqIsCL4wJ3nrfvfZe9g&hl=en&sa=X&ved=0ahUKEwjlxpjm2ZjNAhWMLsAKHX2vAs8Q6AEILzAE#v=onepage&q&f=false
but for this one, the equality hold iff x=-b/2*a and I don't want to put
this as an assumption as I need this lemma in another work. the lemma I'm
lemma
fixes a b c x :: real
assumes "a > 0"
and "⋀x. a*x^2 + b*x + c≥0 "
shows " discrim a b c ≤0"
lemma
fixes a b c x :: real
assumes "a > 0"
and "⋀x. (a*(x)^2 + b*x + c) >0 "
shows " discrim a b c ≤0"
using assms by (metis discriminant_pos_ex less_le not_less)
but for nonnegative equation (the first lemma) I couldn't and way. please
could any one help me in this because I spend alot of time trying to prove
it but unfortunately I failed.
Omar

[Omar Jasim]

Dear Wenda,

now I see where I was stuck. Thank you very much for that, it's really
helpful.

Cheers.

Omar

Post by Wenda Li
Dear Omar,
Here you are,
lemma
fixes a b c x :: real
assumes "a > 0"
and "⋀x. a*x^2 + b*x + c≥0 "
shows "discrim a b c ≤0"
proof -
have "∃x. a*x^2 + b*x + c<0 " when "discrim a b c > 0" "a>0"
proof -
let ?P="λx. a*x^2 + b*x +c"
have "?P (- b / (2*a)) = (4*a*c-b^2)/(4*a)" using ‹a>0›
apply (auto simp add:field_simps)
by algebra
also have "... <0"
using that unfolding discrim_def by (simp add: divide_neg_pos)
finally show ?thesis by blast
qed
then show ?thesis using assms
using not_less by blast
qed
Best,
Wenda

Post by Omar Jasim
Dear all,
Please I need to prove that for any nonnegative quadratic polynomial
equation the discriminant will be nonpositive. I found a proposition in a
https://books.google.co.uk/books?id=TPE0fXGnYtMC&pg=PA81&lpg=PA81&dq=if+f(x)+%3E0+then+b%5E2-4ac%3C0&source=bl&ots=eUAdjNvJpT&sig=KwJxcX_5RqIsCL4wJ3nrfvfZe9g&hl=en&sa=X&ved=0ahUKEwjlxpjm2ZjNAhWMLsAKHX2vAs8Q6AEILzAE#v=onepage&q&f=false
but for this one, the equality hold iff x=-b/2*a and I don't want to put
this as an assumption as I need this lemma in another work. the lemma I'm
lemma
fixes a b c x :: real
assumes "a > 0"
and "⋀x. a*x^2 + b*x + c≥0 "
shows " discrim a b c ≤0"
lemma
fixes a b c x :: real
assumes "a > 0"
and "⋀x. (a*(x)^2 + b*x + c) >0 "
shows " discrim a b c ≤0"
using assms by (metis discriminant_pos_ex less_le not_less)
but for nonnegative equation (the first lemma) I couldn't and way. please
could any one help me in this because I spend alot of time trying to prove
it but unfortunately I failed.
Omar

--
Wenda Li
PhD Candidate
Computer Laboratory
University of Cambridge