Metamath Proof Explorer


Theorem cdj3lem2

Description: Lemma for cdj3i . Value of the first-component function S . (Contributed by NM, 23-May-2005) (New usage is discouraged.)

Ref Expression
Hypotheses cdj3lem2.1 ⊢ A ∈ S ℋ
cdj3lem2.2 ⊢ B ∈ S ℋ
cdj3lem2.3 ⊢ S = x ∈ A + ℋ B ⟼ ι z ∈ A | ∃ w ∈ B x = z + ℎ w
Assertion cdj3lem2 ⊢ C ∈ A ∧ D ∈ B ∧ A ∩ B = 0 ℋ → S ⁡ C + ℎ D = C

Proof

Step Hyp Ref Expression
1 cdj3lem2.1 ⊢ A ∈ S ℋ
2 cdj3lem2.2 ⊢ B ∈ S ℋ
3 cdj3lem2.3 ⊢ S = x ∈ A + ℋ B ⟼ ι z ∈ A | ∃ w ∈ B x = z + ℎ w
4 1 2 shsvai ⊢ C ∈ A ∧ D ∈ B → C + ℎ D ∈ A + ℋ B
5 eqeq1 ⊢ x = C + ℎ D → x = z + ℎ w ↔ C + ℎ D = z + ℎ w
6 5 rexbidv ⊢ x = C + ℎ D → ∃ w ∈ B x = z + ℎ w ↔ ∃ w ∈ B C + ℎ D = z + ℎ w
7 6 riotabidv ⊢ x = C + ℎ D → ι z ∈ A | ∃ w ∈ B x = z + ℎ w = ι z ∈ A | ∃ w ∈ B C + ℎ D = z + ℎ w
8 riotaex ⊢ ι z ∈ A | ∃ w ∈ B C + ℎ D = z + ℎ w ∈ V
9 7 3 8 fvmpt ⊢ C + ℎ D ∈ A + ℋ B → S ⁡ C + ℎ D = ι z ∈ A | ∃ w ∈ B C + ℎ D = z + ℎ w
10 4 9 syl ⊢ C ∈ A ∧ D ∈ B → S ⁡ C + ℎ D = ι z ∈ A | ∃ w ∈ B C + ℎ D = z + ℎ w
11 10 3adant3 ⊢ C ∈ A ∧ D ∈ B ∧ A ∩ B = 0 ℋ → S ⁡ C + ℎ D = ι z ∈ A | ∃ w ∈ B C + ℎ D = z + ℎ w
12 eqid ⊢ C + ℎ D = C + ℎ D
13 oveq2 ⊢ w = D → C + ℎ w = C + ℎ D
14 13 rspceeqv ⊢ D ∈ B ∧ C + ℎ D = C + ℎ D → ∃ w ∈ B C + ℎ D = C + ℎ w
15 12 14 mpan2 ⊢ D ∈ B → ∃ w ∈ B C + ℎ D = C + ℎ w
16 15 3ad2ant2 ⊢ C ∈ A ∧ D ∈ B ∧ A ∩ B = 0 ℋ → ∃ w ∈ B C + ℎ D = C + ℎ w
17 simp1 ⊢ C ∈ A ∧ D ∈ B ∧ A ∩ B = 0 ℋ → C ∈ A
18 1 2 cdjreui ⊢ C + ℎ D ∈ A + ℋ B ∧ A ∩ B = 0 ℋ → ∃! z ∈ A ∃ w ∈ B C + ℎ D = z + ℎ w
19 4 18 stoic3 ⊢ C ∈ A ∧ D ∈ B ∧ A ∩ B = 0 ℋ → ∃! z ∈ A ∃ w ∈ B C + ℎ D = z + ℎ w
20 oveq1 ⊢ z = C → z + ℎ w = C + ℎ w
21 20 eqeq2d ⊢ z = C → C + ℎ D = z + ℎ w ↔ C + ℎ D = C + ℎ w
22 21 rexbidv ⊢ z = C → ∃ w ∈ B C + ℎ D = z + ℎ w ↔ ∃ w ∈ B C + ℎ D = C + ℎ w
23 22 riota2 ⊢ C ∈ A ∧ ∃! z ∈ A ∃ w ∈ B C + ℎ D = z + ℎ w → ∃ w ∈ B C + ℎ D = C + ℎ w ↔ ι z ∈ A | ∃ w ∈ B C + ℎ D = z + ℎ w = C
24 17 19 23 syl2anc ⊢ C ∈ A ∧ D ∈ B ∧ A ∩ B = 0 ℋ → ∃ w ∈ B C + ℎ D = C + ℎ w ↔ ι z ∈ A | ∃ w ∈ B C + ℎ D = z + ℎ w = C
25 16 24 mpbid ⊢ C ∈ A ∧ D ∈ B ∧ A ∩ B = 0 ℋ → ι z ∈ A | ∃ w ∈ B C + ℎ D = z + ℎ w = C
26 11 25 eqtrd ⊢ C ∈ A ∧ D ∈ B ∧ A ∩ B = 0 ℋ → S ⁡ C + ℎ D = C