Metamath Proof Explorer


Theorem imasetpreimafvbijlemfv1

Description: Lemma for imasetpreimafvbij : for a preimage of a value of function F there is an element of the preimage so that the value of the mapping H at this preimage is the function value at this element. (Contributed by AV, 5-Mar-2024)

Ref Expression
Hypotheses fundcmpsurinj.p 𝑃 = { 𝑧 ∣ ∃ 𝑥𝐴 𝑧 = ( 𝐹 “ { ( 𝐹𝑥 ) } ) }
fundcmpsurinj.h 𝐻 = ( 𝑝𝑃 ( 𝐹𝑝 ) )
Assertion imasetpreimafvbijlemfv1 ( ( 𝐹 Fn 𝐴𝑋𝑃 ) → ∃ 𝑦𝑋 ( 𝐻𝑋 ) = ( 𝐹𝑦 ) )

Proof

Step Hyp Ref Expression
1 fundcmpsurinj.p 𝑃 = { 𝑧 ∣ ∃ 𝑥𝐴 𝑧 = ( 𝐹 “ { ( 𝐹𝑥 ) } ) }
2 fundcmpsurinj.h 𝐻 = ( 𝑝𝑃 ( 𝐹𝑝 ) )
3 1 0nelsetpreimafv ( 𝐹 Fn 𝐴 → ∅ ∉ 𝑃 )
4 elnelne2 ( ( 𝑋𝑃 ∧ ∅ ∉ 𝑃 ) → 𝑋 ≠ ∅ )
5 4 expcom ( ∅ ∉ 𝑃 → ( 𝑋𝑃𝑋 ≠ ∅ ) )
6 3 5 syl ( 𝐹 Fn 𝐴 → ( 𝑋𝑃𝑋 ≠ ∅ ) )
7 6 imp ( ( 𝐹 Fn 𝐴𝑋𝑃 ) → 𝑋 ≠ ∅ )
8 simpr ( ( ( 𝐹 Fn 𝐴𝑋𝑃 ) ∧ 𝑦𝑋 ) → 𝑦𝑋 )
9 1 2 imasetpreimafvbijlemfv ( ( 𝐹 Fn 𝐴𝑋𝑃𝑦𝑋 ) → ( 𝐻𝑋 ) = ( 𝐹𝑦 ) )
10 9 3expa ( ( ( 𝐹 Fn 𝐴𝑋𝑃 ) ∧ 𝑦𝑋 ) → ( 𝐻𝑋 ) = ( 𝐹𝑦 ) )
11 8 10 jca ( ( ( 𝐹 Fn 𝐴𝑋𝑃 ) ∧ 𝑦𝑋 ) → ( 𝑦𝑋 ∧ ( 𝐻𝑋 ) = ( 𝐹𝑦 ) ) )
12 11 ex ( ( 𝐹 Fn 𝐴𝑋𝑃 ) → ( 𝑦𝑋 → ( 𝑦𝑋 ∧ ( 𝐻𝑋 ) = ( 𝐹𝑦 ) ) ) )
13 12 eximdv ( ( 𝐹 Fn 𝐴𝑋𝑃 ) → ( ∃ 𝑦 𝑦𝑋 → ∃ 𝑦 ( 𝑦𝑋 ∧ ( 𝐻𝑋 ) = ( 𝐹𝑦 ) ) ) )
14 n0 ( 𝑋 ≠ ∅ ↔ ∃ 𝑦 𝑦𝑋 )
15 df-rex ( ∃ 𝑦𝑋 ( 𝐻𝑋 ) = ( 𝐹𝑦 ) ↔ ∃ 𝑦 ( 𝑦𝑋 ∧ ( 𝐻𝑋 ) = ( 𝐹𝑦 ) ) )
16 13 14 15 3imtr4g ( ( 𝐹 Fn 𝐴𝑋𝑃 ) → ( 𝑋 ≠ ∅ → ∃ 𝑦𝑋 ( 𝐻𝑋 ) = ( 𝐹𝑦 ) ) )
17 7 16 mpd ( ( 𝐹 Fn 𝐴𝑋𝑃 ) → ∃ 𝑦𝑋 ( 𝐻𝑋 ) = ( 𝐹𝑦 ) )