Metamath Proof Explorer


Theorem i1fmulclem

Description: Decompose the preimage of a constant times a function. (Contributed by Mario Carneiro, 25-Jun-2014)

Ref Expression
Hypotheses i1fmulc.2 ⊢ φ → F ∈ dom ⁡ ∫ 1
i1fmulc.3 ⊢ φ → A ∈ ℝ
Assertion i1fmulclem ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ → ℝ × A × f F -1 B = F -1 B A

Proof

Step Hyp Ref Expression
1 i1fmulc.2 ⊢ φ → F ∈ dom ⁡ ∫ 1
2 i1fmulc.3 ⊢ φ → A ∈ ℝ
3 reex ⊢ ℝ ∈ V
4 3 a1i ⊢ φ → ℝ ∈ V
5 i1ff ⊢ F ∈ dom ⁡ ∫ 1 → F : ℝ ⟶ ℝ
6 1 5 syl ⊢ φ → F : ℝ ⟶ ℝ
7 6 ffnd ⊢ φ → F Fn ℝ
8 eqidd ⊢ φ ∧ z ∈ ℝ → F ⁡ z = F ⁡ z
9 4 2 7 8 ofc1 ⊢ φ ∧ z ∈ ℝ → ℝ × A × f F ⁡ z = A ⁢ F ⁡ z
10 9 ad4ant14 ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ ∧ z ∈ ℝ → ℝ × A × f F ⁡ z = A ⁢ F ⁡ z
11 10 eqeq1d ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ ∧ z ∈ ℝ → ℝ × A × f F ⁡ z = B ↔ A ⁢ F ⁡ z = B
12 eqcom ⊢ F ⁡ z = B A ↔ B A = F ⁡ z
13 simplr ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ ∧ z ∈ ℝ → B ∈ ℝ
14 13 recnd ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ ∧ z ∈ ℝ → B ∈ ℂ
15 2 ad3antrrr ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ ∧ z ∈ ℝ → A ∈ ℝ
16 15 recnd ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ ∧ z ∈ ℝ → A ∈ ℂ
17 6 ad2antrr ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ → F : ℝ ⟶ ℝ
18 17 ffvelcdmda ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ ∧ z ∈ ℝ → F ⁡ z ∈ ℝ
19 18 recnd ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ ∧ z ∈ ℝ → F ⁡ z ∈ ℂ
20 simpllr ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ ∧ z ∈ ℝ → A ≠ 0
21 14 16 19 20 divmuld ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ ∧ z ∈ ℝ → B A = F ⁡ z ↔ A ⁢ F ⁡ z = B
22 12 21 bitrid ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ ∧ z ∈ ℝ → F ⁡ z = B A ↔ A ⁢ F ⁡ z = B
23 11 22 bitr4d ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ ∧ z ∈ ℝ → ℝ × A × f F ⁡ z = B ↔ F ⁡ z = B A
24 23 pm5.32da ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ → z ∈ ℝ ∧ ℝ × A × f F ⁡ z = B ↔ z ∈ ℝ ∧ F ⁡ z = B A
25 remulcl ⊢ x ∈ ℝ ∧ y ∈ ℝ → x ⁢ y ∈ ℝ
26 25 adantl ⊢ φ ∧ x ∈ ℝ ∧ y ∈ ℝ → x ⁢ y ∈ ℝ
27 fconstg ⊢ A ∈ ℝ → ℝ × A : ℝ ⟶ A
28 2 27 syl ⊢ φ → ℝ × A : ℝ ⟶ A
29 2 snssd ⊢ φ → A ⊆ ℝ
30 28 29 fssd ⊢ φ → ℝ × A : ℝ ⟶ ℝ
31 inidm ⊢ ℝ ∩ ℝ = ℝ
32 26 30 6 4 4 31 off ⊢ φ → ℝ × A × f F : ℝ ⟶ ℝ
33 32 ad2antrr ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ → ℝ × A × f F : ℝ ⟶ ℝ
34 33 ffnd ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ → ℝ × A × f F Fn ℝ
35 fniniseg ⊢ ℝ × A × f F Fn ℝ → z ∈ ℝ × A × f F -1 B ↔ z ∈ ℝ ∧ ℝ × A × f F ⁡ z = B
36 34 35 syl ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ → z ∈ ℝ × A × f F -1 B ↔ z ∈ ℝ ∧ ℝ × A × f F ⁡ z = B
37 17 ffnd ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ → F Fn ℝ
38 fniniseg ⊢ F Fn ℝ → z ∈ F -1 B A ↔ z ∈ ℝ ∧ F ⁡ z = B A
39 37 38 syl ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ → z ∈ F -1 B A ↔ z ∈ ℝ ∧ F ⁡ z = B A
40 24 36 39 3bitr4d ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ → z ∈ ℝ × A × f F -1 B ↔ z ∈ F -1 B A
41 40 eqrdv ⊢ φ ∧ A ≠ 0 ∧ B ∈ ℝ → ℝ × A × f F -1 B = F -1 B A