Metamath Proof Explorer


Theorem txindislem

Description: Lemma for txindis . (Contributed by Mario Carneiro, 14-Aug-2015)

Ref Expression
Assertion txindislem ⊢ I ⁡ A × I ⁡ B = I ⁡ A × B

Proof

Step Hyp Ref Expression
1 0xp ⊢ ∅ × I ⁡ B = ∅
2 fvprc ⊢ ¬ A ∈ V → I ⁡ A = ∅
3 2 xpeq1d ⊢ ¬ A ∈ V → I ⁡ A × I ⁡ B = ∅ × I ⁡ B
4 simpr ⊢ ¬ A ∈ V ∧ B = ∅ → B = ∅
5 4 xpeq2d ⊢ ¬ A ∈ V ∧ B = ∅ → A × B = A × ∅
6 xp0 ⊢ A × ∅ = ∅
7 5 6 eqtrdi ⊢ ¬ A ∈ V ∧ B = ∅ → A × B = ∅
8 7 fveq2d ⊢ ¬ A ∈ V ∧ B = ∅ → I ⁡ A × B = I ⁡ ∅
9 0ex ⊢ ∅ ∈ V
10 fvi ⊢ ∅ ∈ V → I ⁡ ∅ = ∅
11 9 10 ax-mp ⊢ I ⁡ ∅ = ∅
12 8 11 eqtrdi ⊢ ¬ A ∈ V ∧ B = ∅ → I ⁡ A × B = ∅
13 dmexg ⊢ A × B ∈ V → dom ⁡ A × B ∈ V
14 dmxp ⊢ B ≠ ∅ → dom ⁡ A × B = A
15 14 eleq1d ⊢ B ≠ ∅ → dom ⁡ A × B ∈ V ↔ A ∈ V
16 13 15 imbitrid ⊢ B ≠ ∅ → A × B ∈ V → A ∈ V
17 16 con3d ⊢ B ≠ ∅ → ¬ A ∈ V → ¬ A × B ∈ V
18 17 impcom ⊢ ¬ A ∈ V ∧ B ≠ ∅ → ¬ A × B ∈ V
19 fvprc ⊢ ¬ A × B ∈ V → I ⁡ A × B = ∅
20 18 19 syl ⊢ ¬ A ∈ V ∧ B ≠ ∅ → I ⁡ A × B = ∅
21 12 20 pm2.61dane ⊢ ¬ A ∈ V → I ⁡ A × B = ∅
22 1 3 21 3eqtr4a ⊢ ¬ A ∈ V → I ⁡ A × I ⁡ B = I ⁡ A × B
23 xp0 ⊢ I ⁡ A × ∅ = ∅
24 fvprc ⊢ ¬ B ∈ V → I ⁡ B = ∅
25 24 xpeq2d ⊢ ¬ B ∈ V → I ⁡ A × I ⁡ B = I ⁡ A × ∅
26 simpr ⊢ ¬ B ∈ V ∧ A = ∅ → A = ∅
27 26 xpeq1d ⊢ ¬ B ∈ V ∧ A = ∅ → A × B = ∅ × B
28 0xp ⊢ ∅ × B = ∅
29 27 28 eqtrdi ⊢ ¬ B ∈ V ∧ A = ∅ → A × B = ∅
30 29 fveq2d ⊢ ¬ B ∈ V ∧ A = ∅ → I ⁡ A × B = I ⁡ ∅
31 30 11 eqtrdi ⊢ ¬ B ∈ V ∧ A = ∅ → I ⁡ A × B = ∅
32 rnexg ⊢ A × B ∈ V → ran ⁡ A × B ∈ V
33 rnxp ⊢ A ≠ ∅ → ran ⁡ A × B = B
34 33 eleq1d ⊢ A ≠ ∅ → ran ⁡ A × B ∈ V ↔ B ∈ V
35 32 34 imbitrid ⊢ A ≠ ∅ → A × B ∈ V → B ∈ V
36 35 con3d ⊢ A ≠ ∅ → ¬ B ∈ V → ¬ A × B ∈ V
37 36 impcom ⊢ ¬ B ∈ V ∧ A ≠ ∅ → ¬ A × B ∈ V
38 37 19 syl ⊢ ¬ B ∈ V ∧ A ≠ ∅ → I ⁡ A × B = ∅
39 31 38 pm2.61dane ⊢ ¬ B ∈ V → I ⁡ A × B = ∅
40 23 25 39 3eqtr4a ⊢ ¬ B ∈ V → I ⁡ A × I ⁡ B = I ⁡ A × B
41 fvi ⊢ A ∈ V → I ⁡ A = A
42 fvi ⊢ B ∈ V → I ⁡ B = B
43 xpeq12 ⊢ I ⁡ A = A ∧ I ⁡ B = B → I ⁡ A × I ⁡ B = A × B
44 41 42 43 syl2an ⊢ A ∈ V ∧ B ∈ V → I ⁡ A × I ⁡ B = A × B
45 xpexg ⊢ A ∈ V ∧ B ∈ V → A × B ∈ V
46 fvi ⊢ A × B ∈ V → I ⁡ A × B = A × B
47 45 46 syl ⊢ A ∈ V ∧ B ∈ V → I ⁡ A × B = A × B
48 44 47 eqtr4d ⊢ A ∈ V ∧ B ∈ V → I ⁡ A × I ⁡ B = I ⁡ A × B
49 22 40 48 ecase ⊢ I ⁡ A × I ⁡ B = I ⁡ A × B