Metamath Proof Explorer


Theorem fundcmpsurinjlem1

Description: Lemma 1 for fundcmpsurinj . (Contributed by AV, 4-Mar-2024)

Ref Expression
Hypotheses fundcmpsurinj.p ⊢ P = z | ∃ x ∈ A z = F -1 F ⁡ x
fundcmpsurinj.g ⊢ G = x ∈ A ⟼ F -1 F ⁡ x
Assertion fundcmpsurinjlem1 ⊢ ran ⁡ G = P

Proof

Step Hyp Ref Expression
1 fundcmpsurinj.p ⊢ P = z | ∃ x ∈ A z = F -1 F ⁡ x
2 fundcmpsurinj.g ⊢ G = x ∈ A ⟼ F -1 F ⁡ x
3 2 rnmpt ⊢ ran ⁡ G = z | ∃ x ∈ A z = F -1 F ⁡ x
4 3 1 eqtr4i ⊢ ran ⁡ G = P