Metamath Proof Explorer


Theorem cdj3lem2a

Description: Lemma for cdj3i . Closure of the first-component function S . (Contributed by NM, 25-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 cdj3lem2a ⊢ C ∈ A + ℋ B ∧ A ∩ B = 0 ℋ → S ⁡ C ∈ A

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 shseli ⊢ C ∈ A + ℋ B ↔ ∃ v ∈ A ∃ u ∈ B C = v + ℎ u
5 1 2 3 cdj3lem2 ⊢ v ∈ A ∧ u ∈ B ∧ A ∩ B = 0 ℋ → S ⁡ v + ℎ u = v
6 simp1 ⊢ v ∈ A ∧ u ∈ B ∧ A ∩ B = 0 ℋ → v ∈ A
7 5 6 eqeltrd ⊢ v ∈ A ∧ u ∈ B ∧ A ∩ B = 0 ℋ → S ⁡ v + ℎ u ∈ A
8 7 3expa ⊢ v ∈ A ∧ u ∈ B ∧ A ∩ B = 0 ℋ → S ⁡ v + ℎ u ∈ A
9 fveq2 ⊢ C = v + ℎ u → S ⁡ C = S ⁡ v + ℎ u
10 9 eleq1d ⊢ C = v + ℎ u → S ⁡ C ∈ A ↔ S ⁡ v + ℎ u ∈ A
11 8 10 imbitrrid ⊢ C = v + ℎ u → v ∈ A ∧ u ∈ B ∧ A ∩ B = 0 ℋ → S ⁡ C ∈ A
12 11 expd ⊢ C = v + ℎ u → v ∈ A ∧ u ∈ B → A ∩ B = 0 ℋ → S ⁡ C ∈ A
13 12 com13 ⊢ A ∩ B = 0 ℋ → v ∈ A ∧ u ∈ B → C = v + ℎ u → S ⁡ C ∈ A
14 13 rexlimdvv ⊢ A ∩ B = 0 ℋ → ∃ v ∈ A ∃ u ∈ B C = v + ℎ u → S ⁡ C ∈ A
15 4 14 biimtrid ⊢ A ∩ B = 0 ℋ → C ∈ A + ℋ B → S ⁡ C ∈ A
16 15 impcom ⊢ C ∈ A + ℋ B ∧ A ∩ B = 0 ℋ → S ⁡ C ∈ A