Distribution law for the inverse image operation
with respect to intersection.
Proof sketch
The inclusion f−1[A ∩ B] ⊆ f−1[A] ∩ f−1[B]
is universally valid for any relation f. Thus it suffices to show the
converse inclusion
f−1[A] ∩ f−1[B] ⊆ f−1[A ∩ B].
So let x ∈ f−1[A] ∩ f−1[B], thus
x ∈ f−1[A] as well as x ∈ f−1[B].
In this respect, a ∈ A with (x, a) ∈ f and b ∈ B with (x, b) ∈ f exist.
However, because f is a function, a = b follows from (x, a) ∈ f and
(x, b) ∈ f, since a functions's argument has a unique value. Accordingly,
a ∈ B holds, meaning that a ∈ A ∩ B with (x, a) ∈ f, which
witnesses that x ∈ f−1[A ∩ B]. q.e.d.
Proof 01. 1 ⊢ function f, hypo. 02. 2 ⊢ x ∈ inv_img f A ∩ inv_img f B, hypo. 03. 2 ⊢ x ∈ inv_img f A, intersection_eliml 2. 04. 2 ⊢ x ∈ inv_img f B, intersection_elimr 2. 05. 2 ⊢ ∃a. a ∈ A ∧ (x, a) ∈ f, inv_img_elim 3. 06. 2 ⊢ ∃b. b ∈ B ∧ (x, b) ∈ f, inv_img_elim 4. 07. 7 ⊢ a ∈ A ∧ (x, a) ∈ f, hypo. 08. 8 ⊢ b ∈ B ∧ (x, b) ∈ f, hypo. 09. 7 ⊢ a ∈ A, conj_eliml 7. 10. 8 ⊢ b ∈ B, conj_eliml 8. 11. 7 ⊢ (x, a) ∈ f, conj_elimr 7. 12. 8 ⊢ (x, b) ∈ f, conj_elimr 8. 13. 1, 7, 8 ⊢ a = b, fn_unique_value 1 11 12. 14. 1, 7, 8 ⊢ a ∈ B, eq_subst_rev 13 10, P t ↔ t ∈ B. 15. 1, 7, 8 ⊢ a ∈ A ∩ B, intersection_intro 9 14. 16. 1, 7, 8 ⊢ x ∈ inv_img f (A ∩ B), inv_img_intro 15 11. 17. 1, 2, 7 ⊢ x ∈ inv_img f (A ∩ B), ex_elim 6 16. 18. 1, 2 ⊢ x ∈ inv_img f (A ∩ B), ex_elim 5 17. 19. 1 ⊢ x ∈ inv_img f A ∩ inv_img f B → x ∈ inv_img f (A ∩ B), subj_intro 18. 20. 1 ⊢ ∀x. x ∈ inv_img f A ∩ inv_img f B → x ∈ inv_img f (A ∩ B), uq_intro 19. 21. 1 ⊢ inv_img f A ∩ inv_img f B ⊆ inv_img f (A ∩ B), incl_intro 20. 22. ⊢ inv_img f (A ∩ B) ⊆ inv_img f A ∩ inv_img f B, inv_img_dist_inter. 23. 1 ⊢ inv_img f (A ∩ B) = inv_img f A ∩ inv_img f B, incl_antisym 22 21. fn_inv_img_dist_inter. ⊢ function f → inv_img f (A ∩ B) = inv_img f A ∩ inv_img f B, subj_intro 23.
Dependencies
The given proof depends on seven axioms:
comp, efq, eq_refl, eq_subst, ext, lem, subset.