Metamath Proof Explorer


Theorem xpsrnbas

Description: The indexed structure product that appears in xpsval has the same base as the target of the function F . (Contributed by Mario Carneiro, 15-Aug-2015) (Revised by Jim Kingdon, 25-Sep-2023)

Ref Expression
Hypotheses xpsval.t ⊢ T = R × 𝑠 S
xpsval.x ⊢ X = Base R
xpsval.y ⊢ Y = Base S
xpsval.1 ⊢ φ → R ∈ V
xpsval.2 ⊢ φ → S ∈ W
xpsval.f ⊢ F = x ∈ X , y ∈ Y ⟼ ∅ x 1 𝑜 y
xpsval.k ⊢ G = Scalar ⁡ R
xpsval.u ⊢ U = G ⨉ 𝑠 ∅ R 1 𝑜 S
Assertion xpsrnbas ⊢ φ → ran ⁡ F = Base U

Proof

Step Hyp Ref Expression
1 xpsval.t ⊢ T = R × 𝑠 S
2 xpsval.x ⊢ X = Base R
3 xpsval.y ⊢ Y = Base S
4 xpsval.1 ⊢ φ → R ∈ V
5 xpsval.2 ⊢ φ → S ∈ W
6 xpsval.f ⊢ F = x ∈ X , y ∈ Y ⟼ ∅ x 1 𝑜 y
7 xpsval.k ⊢ G = Scalar ⁡ R
8 xpsval.u ⊢ U = G ⨉ 𝑠 ∅ R 1 𝑜 S
9 eqid ⊢ Base U = Base U
10 7 fvexi ⊢ G ∈ V
11 10 a1i ⊢ φ → G ∈ V
12 2on ⊢ 2 𝑜 ∈ On
13 12 a1i ⊢ φ → 2 𝑜 ∈ On
14 fnpr2o ⊢ R ∈ V ∧ S ∈ W → ∅ R 1 𝑜 S Fn 2 𝑜
15 4 5 14 syl2anc ⊢ φ → ∅ R 1 𝑜 S Fn 2 𝑜
16 8 9 11 13 15 prdsbas2 ⊢ φ → Base U = ⨉ k ∈ 2 𝑜 Base ∅ R 1 𝑜 S ⁡ k
17 fvprif ⊢ R ∈ V ∧ S ∈ W ∧ k ∈ 2 𝑜 → ∅ R 1 𝑜 S ⁡ k = if k = ∅ R S
18 17 3expia ⊢ R ∈ V ∧ S ∈ W → k ∈ 2 𝑜 → ∅ R 1 𝑜 S ⁡ k = if k = ∅ R S
19 4 5 18 syl2anc ⊢ φ → k ∈ 2 𝑜 → ∅ R 1 𝑜 S ⁡ k = if k = ∅ R S
20 19 imp ⊢ φ ∧ k ∈ 2 𝑜 → ∅ R 1 𝑜 S ⁡ k = if k = ∅ R S
21 20 fveq2d ⊢ φ ∧ k ∈ 2 𝑜 → Base ∅ R 1 𝑜 S ⁡ k = Base if k = ∅ R S
22 ifeq12 ⊢ X = Base R ∧ Y = Base S → if k = ∅ X Y = if k = ∅ Base R Base S
23 2 3 22 mp2an ⊢ if k = ∅ X Y = if k = ∅ Base R Base S
24 fvif ⊢ Base if k = ∅ R S = if k = ∅ Base R Base S
25 23 24 eqtr4i ⊢ if k = ∅ X Y = Base if k = ∅ R S
26 21 25 eqtr4di ⊢ φ ∧ k ∈ 2 𝑜 → Base ∅ R 1 𝑜 S ⁡ k = if k = ∅ X Y
27 26 ixpeq2dva ⊢ φ → ⨉ k ∈ 2 𝑜 Base ∅ R 1 𝑜 S ⁡ k = ⨉ k ∈ 2 𝑜 if k = ∅ X Y
28 6 xpsfrn ⊢ ran ⁡ F = ⨉ k ∈ 2 𝑜 if k = ∅ X Y
29 27 28 eqtr4di ⊢ φ → ⨉ k ∈ 2 𝑜 Base ∅ R 1 𝑜 S ⁡ k = ran ⁡ F
30 16 29 eqtr2d ⊢ φ → ran ⁡ F = Base U