Metamath Proof Explorer


Theorem fliftcnv

Description: Converse of the relation F . (Contributed by Mario Carneiro, 23-Dec-2016)

Ref Expression
Hypotheses flift.1 ⊢ F = ran ⁡ x ∈ X ⟼ A B
flift.2 ⊢ φ ∧ x ∈ X → A ∈ R
flift.3 ⊢ φ ∧ x ∈ X → B ∈ S
Assertion fliftcnv ⊢ φ → F -1 = ran ⁡ x ∈ X ⟼ B A

Proof

Step Hyp Ref Expression
1 flift.1 ⊢ F = ran ⁡ x ∈ X ⟼ A B
2 flift.2 ⊢ φ ∧ x ∈ X → A ∈ R
3 flift.3 ⊢ φ ∧ x ∈ X → B ∈ S
4 eqid ⊢ ran ⁡ x ∈ X ⟼ B A = ran ⁡ x ∈ X ⟼ B A
5 4 3 2 fliftrel ⊢ φ → ran ⁡ x ∈ X ⟼ B A ⊆ S × R
6 relxp ⊢ Rel ⁡ S × R
7 relss ⊢ ran ⁡ x ∈ X ⟼ B A ⊆ S × R → Rel ⁡ S × R → Rel ⁡ ran ⁡ x ∈ X ⟼ B A
8 5 6 7 mpisyl ⊢ φ → Rel ⁡ ran ⁡ x ∈ X ⟼ B A
9 relcnv ⊢ Rel ⁡ F -1
10 8 9 jctil ⊢ φ → Rel ⁡ F -1 ∧ Rel ⁡ ran ⁡ x ∈ X ⟼ B A
11 1 2 3 fliftel ⊢ φ → z F y ↔ ∃ x ∈ X z = A ∧ y = B
12 vex ⊢ y ∈ V
13 vex ⊢ z ∈ V
14 12 13 brcnv ⊢ y F -1 z ↔ z F y
15 ancom ⊢ y = B ∧ z = A ↔ z = A ∧ y = B
16 15 rexbii ⊢ ∃ x ∈ X y = B ∧ z = A ↔ ∃ x ∈ X z = A ∧ y = B
17 11 14 16 3bitr4g ⊢ φ → y F -1 z ↔ ∃ x ∈ X y = B ∧ z = A
18 4 3 2 fliftel ⊢ φ → y ran ⁡ x ∈ X ⟼ B A z ↔ ∃ x ∈ X y = B ∧ z = A
19 17 18 bitr4d ⊢ φ → y F -1 z ↔ y ran ⁡ x ∈ X ⟼ B A z
20 df-br ⊢ y F -1 z ↔ y z ∈ F -1
21 df-br ⊢ y ran ⁡ x ∈ X ⟼ B A z ↔ y z ∈ ran ⁡ x ∈ X ⟼ B A
22 19 20 21 3bitr3g ⊢ φ → y z ∈ F -1 ↔ y z ∈ ran ⁡ x ∈ X ⟼ B A
23 22 eqrelrdv2 ⊢ Rel ⁡ F -1 ∧ Rel ⁡ ran ⁡ x ∈ X ⟼ B A ∧ φ → F -1 = ran ⁡ x ∈ X ⟼ B A
24 10 23 mpancom ⊢ φ → F -1 = ran ⁡ x ∈ X ⟼ B A