Metamath Proof Explorer


Theorem fundcmpsurinjlem1

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

Ref Expression
Hypotheses fundcmpsurinj.p 𝑃 = { 𝑧 ∣ ∃ 𝑥𝐴 𝑧 = ( 𝐹 “ { ( 𝐹𝑥 ) } ) }
fundcmpsurinj.g 𝐺 = ( 𝑥𝐴 ↦ ( 𝐹 “ { ( 𝐹𝑥 ) } ) )
Assertion fundcmpsurinjlem1 ran 𝐺 = 𝑃

Proof

Step Hyp Ref Expression
1 fundcmpsurinj.p 𝑃 = { 𝑧 ∣ ∃ 𝑥𝐴 𝑧 = ( 𝐹 “ { ( 𝐹𝑥 ) } ) }
2 fundcmpsurinj.g 𝐺 = ( 𝑥𝐴 ↦ ( 𝐹 “ { ( 𝐹𝑥 ) } ) )
3 2 rnmpt ran 𝐺 = { 𝑧 ∣ ∃ 𝑥𝐴 𝑧 = ( 𝐹 “ { ( 𝐹𝑥 ) } ) }
4 3 1 eqtr4i ran 𝐺 = 𝑃