Metamath Proof Explorer


Theorem bj-projeq

Description: Substitution property for Proj . (Contributed by BJ, 6-Apr-2019)

Ref Expression
Assertion bj-projeq ⊢ A = C → B = D → A Proj B = C Proj D

Proof

Step Hyp Ref Expression
1 simpr ⊢ A = C ∧ B = D → B = D
2 simpl ⊢ A = C ∧ B = D → A = C
3 2 sneqd ⊢ A = C ∧ B = D → A = C
4 1 3 imaeq12d ⊢ A = C ∧ B = D → B A = D C
5 4 eleq2d ⊢ A = C ∧ B = D → x ∈ B A ↔ x ∈ D C
6 5 abbidv ⊢ A = C ∧ B = D → x | x ∈ B A = x | x ∈ D C
7 df-bj-proj ⊢ A Proj B = x | x ∈ B A
8 df-bj-proj ⊢ C Proj D = x | x ∈ D C
9 6 7 8 3eqtr4g ⊢ A = C ∧ B = D → A Proj B = C Proj D
10 9 ex ⊢ A = C → B = D → A Proj B = C Proj D