Metamath Proof Explorer


Theorem 5oalem4

Description: Lemma for orthoarguesian law 5OA. (Contributed by NM, 2-Apr-2000) (New usage is discouraged.)

Ref Expression
Hypotheses 5oalem3.1 ⊢ A ∈ S ℋ
5oalem3.2 ⊢ B ∈ S ℋ
5oalem3.3 ⊢ C ∈ S ℋ
5oalem3.4 ⊢ D ∈ S ℋ
5oalem3.5 ⊢ F ∈ S ℋ
5oalem3.6 ⊢ G ∈ S ℋ
Assertion 5oalem4 ⊢ x ∈ A ∧ y ∈ B ∧ z ∈ C ∧ w ∈ D ∧ f ∈ F ∧ g ∈ G ∧ x + ℎ y = f + ℎ g ∧ z + ℎ w = f + ℎ g → x - ℎ z ∈ A + ℋ C ∩ B + ℋ D ∩ A + ℋ F ∩ B + ℋ G + ℋ C + ℋ F ∩ D + ℋ G

Proof

Step Hyp Ref Expression
1 5oalem3.1 ⊢ A ∈ S ℋ
2 5oalem3.2 ⊢ B ∈ S ℋ
3 5oalem3.3 ⊢ C ∈ S ℋ
4 5oalem3.4 ⊢ D ∈ S ℋ
5 5oalem3.5 ⊢ F ∈ S ℋ
6 5oalem3.6 ⊢ G ∈ S ℋ
7 eqtr3 ⊢ x + ℎ y = f + ℎ g ∧ z + ℎ w = f + ℎ g → x + ℎ y = z + ℎ w
8 1 2 3 4 5oalem2 ⊢ x ∈ A ∧ y ∈ B ∧ z ∈ C ∧ w ∈ D ∧ x + ℎ y = z + ℎ w → x - ℎ z ∈ A + ℋ C ∩ B + ℋ D
9 7 8 sylan2 ⊢ x ∈ A ∧ y ∈ B ∧ z ∈ C ∧ w ∈ D ∧ x + ℎ y = f + ℎ g ∧ z + ℎ w = f + ℎ g → x - ℎ z ∈ A + ℋ C ∩ B + ℋ D
10 9 adantlr ⊢ x ∈ A ∧ y ∈ B ∧ z ∈ C ∧ w ∈ D ∧ f ∈ F ∧ g ∈ G ∧ x + ℎ y = f + ℎ g ∧ z + ℎ w = f + ℎ g → x - ℎ z ∈ A + ℋ C ∩ B + ℋ D
11 1 2 3 4 5 6 5oalem3 ⊢ x ∈ A ∧ y ∈ B ∧ z ∈ C ∧ w ∈ D ∧ f ∈ F ∧ g ∈ G ∧ x + ℎ y = f + ℎ g ∧ z + ℎ w = f + ℎ g → x - ℎ z ∈ A + ℋ F ∩ B + ℋ G + ℋ C + ℋ F ∩ D + ℋ G
12 10 11 elind ⊢ x ∈ A ∧ y ∈ B ∧ z ∈ C ∧ w ∈ D ∧ f ∈ F ∧ g ∈ G ∧ x + ℎ y = f + ℎ g ∧ z + ℎ w = f + ℎ g → x - ℎ z ∈ A + ℋ C ∩ B + ℋ D ∩ A + ℋ F ∩ B + ℋ G + ℋ C + ℋ F ∩ D + ℋ G