Metamath Proof Explorer


Theorem ofrn2

Description: The range of the function operation. (Contributed by Thierry Arnoux, 21-Mar-2017)

Ref Expression
Hypotheses ofrn.1 ⊢ φ → F : A ⟶ B
ofrn.2 ⊢ φ → G : A ⟶ B
ofrn.3 ⊢ φ → + ˙ : B × B ⟶ C
ofrn.4 ⊢ φ → A ∈ V
Assertion ofrn2 ⊢ φ → ran ⁡ F + ˙ f G ⊆ + ˙ ran ⁡ F × ran ⁡ G

Proof

Step Hyp Ref Expression
1 ofrn.1 ⊢ φ → F : A ⟶ B
2 ofrn.2 ⊢ φ → G : A ⟶ B
3 ofrn.3 ⊢ φ → + ˙ : B × B ⟶ C
4 ofrn.4 ⊢ φ → A ∈ V
5 1 ffnd ⊢ φ → F Fn A
6 simprl ⊢ φ ∧ a ∈ A ∧ z = F ⁡ a + ˙ G ⁡ a → a ∈ A
7 fnfvelrn ⊢ F Fn A ∧ a ∈ A → F ⁡ a ∈ ran ⁡ F
8 5 6 7 syl2an2r ⊢ φ ∧ a ∈ A ∧ z = F ⁡ a + ˙ G ⁡ a → F ⁡ a ∈ ran ⁡ F
9 2 ffnd ⊢ φ → G Fn A
10 fnfvelrn ⊢ G Fn A ∧ a ∈ A → G ⁡ a ∈ ran ⁡ G
11 9 6 10 syl2an2r ⊢ φ ∧ a ∈ A ∧ z = F ⁡ a + ˙ G ⁡ a → G ⁡ a ∈ ran ⁡ G
12 simprr ⊢ φ ∧ a ∈ A ∧ z = F ⁡ a + ˙ G ⁡ a → z = F ⁡ a + ˙ G ⁡ a
13 rspceov ⊢ F ⁡ a ∈ ran ⁡ F ∧ G ⁡ a ∈ ran ⁡ G ∧ z = F ⁡ a + ˙ G ⁡ a → ∃ x ∈ ran ⁡ F ∃ y ∈ ran ⁡ G z = x + ˙ y
14 8 11 12 13 syl3anc ⊢ φ ∧ a ∈ A ∧ z = F ⁡ a + ˙ G ⁡ a → ∃ x ∈ ran ⁡ F ∃ y ∈ ran ⁡ G z = x + ˙ y
15 14 rexlimdvaa ⊢ φ → ∃ a ∈ A z = F ⁡ a + ˙ G ⁡ a → ∃ x ∈ ran ⁡ F ∃ y ∈ ran ⁡ G z = x + ˙ y
16 15 ss2abdv ⊢ φ → z | ∃ a ∈ A z = F ⁡ a + ˙ G ⁡ a ⊆ z | ∃ x ∈ ran ⁡ F ∃ y ∈ ran ⁡ G z = x + ˙ y
17 inidm ⊢ A ∩ A = A
18 eqidd ⊢ φ ∧ a ∈ A → F ⁡ a = F ⁡ a
19 eqidd ⊢ φ ∧ a ∈ A → G ⁡ a = G ⁡ a
20 5 9 4 4 17 18 19 offval ⊢ φ → F + ˙ f G = a ∈ A ⟼ F ⁡ a + ˙ G ⁡ a
21 20 rneqd ⊢ φ → ran ⁡ F + ˙ f G = ran ⁡ a ∈ A ⟼ F ⁡ a + ˙ G ⁡ a
22 eqid ⊢ a ∈ A ⟼ F ⁡ a + ˙ G ⁡ a = a ∈ A ⟼ F ⁡ a + ˙ G ⁡ a
23 22 rnmpt ⊢ ran ⁡ a ∈ A ⟼ F ⁡ a + ˙ G ⁡ a = z | ∃ a ∈ A z = F ⁡ a + ˙ G ⁡ a
24 21 23 eqtrdi ⊢ φ → ran ⁡ F + ˙ f G = z | ∃ a ∈ A z = F ⁡ a + ˙ G ⁡ a
25 3 ffnd ⊢ φ → + ˙ Fn B × B
26 1 frnd ⊢ φ → ran ⁡ F ⊆ B
27 2 frnd ⊢ φ → ran ⁡ G ⊆ B
28 xpss12 ⊢ ran ⁡ F ⊆ B ∧ ran ⁡ G ⊆ B → ran ⁡ F × ran ⁡ G ⊆ B × B
29 26 27 28 syl2anc ⊢ φ → ran ⁡ F × ran ⁡ G ⊆ B × B
30 ovelimab ⊢ + ˙ Fn B × B ∧ ran ⁡ F × ran ⁡ G ⊆ B × B → z ∈ + ˙ ran ⁡ F × ran ⁡ G ↔ ∃ x ∈ ran ⁡ F ∃ y ∈ ran ⁡ G z = x + ˙ y
31 25 29 30 syl2anc ⊢ φ → z ∈ + ˙ ran ⁡ F × ran ⁡ G ↔ ∃ x ∈ ran ⁡ F ∃ y ∈ ran ⁡ G z = x + ˙ y
32 31 eqabdv ⊢ φ → + ˙ ran ⁡ F × ran ⁡ G = z | ∃ x ∈ ran ⁡ F ∃ y ∈ ran ⁡ G z = x + ˙ y
33 16 24 32 3sstr4d ⊢ φ → ran ⁡ F + ˙ f G ⊆ + ˙ ran ⁡ F × ran ⁡ G