Metamath Proof Explorer


Theorem imaelsetpreimafv

Description: The image of an element of the preimage of a function value is the singleton consisting of the function value at one of its elements. (Contributed by AV, 5-Mar-2024)

Ref Expression
Hypothesis setpreimafvex.p ⊢ P = z | ∃ x ∈ A z = F -1 F ⁡ x
Assertion imaelsetpreimafv ⊢ F Fn A ∧ S ∈ P ∧ X ∈ S → F S = F ⁡ X

Proof

Step Hyp Ref Expression
1 setpreimafvex.p ⊢ P = z | ∃ x ∈ A z = F -1 F ⁡ x
2 1 fvelsetpreimafv ⊢ F Fn A ∧ S ∈ P → ∃ x ∈ S S = F -1 F ⁡ x
3 fveq2 ⊢ y = x → F ⁡ y = F ⁡ x
4 3 sneqd ⊢ y = x → F ⁡ y = F ⁡ x
5 4 imaeq2d ⊢ y = x → F -1 F ⁡ y = F -1 F ⁡ x
6 5 eqeq2d ⊢ y = x → S = F -1 F ⁡ y ↔ S = F -1 F ⁡ x
7 6 cbvrexvw ⊢ ∃ y ∈ S S = F -1 F ⁡ y ↔ ∃ x ∈ S S = F -1 F ⁡ x
8 2 7 sylibr ⊢ F Fn A ∧ S ∈ P → ∃ y ∈ S S = F -1 F ⁡ y
9 8 3adant3 ⊢ F Fn A ∧ S ∈ P ∧ X ∈ S → ∃ y ∈ S S = F -1 F ⁡ y
10 imaeq2 ⊢ S = F -1 F ⁡ y → F S = F F -1 F ⁡ y
11 10 3ad2ant3 ⊢ F Fn A ∧ S ∈ P ∧ X ∈ S ∧ y ∈ S ∧ S = F -1 F ⁡ y → F S = F F -1 F ⁡ y
12 fnfun ⊢ F Fn A → Fun ⁡ F
13 funimacnv ⊢ Fun ⁡ F → F F -1 F ⁡ y = F ⁡ y ∩ ran ⁡ F
14 12 13 syl ⊢ F Fn A → F F -1 F ⁡ y = F ⁡ y ∩ ran ⁡ F
15 14 3ad2ant1 ⊢ F Fn A ∧ S ∈ P ∧ X ∈ S → F F -1 F ⁡ y = F ⁡ y ∩ ran ⁡ F
16 15 3ad2ant1 ⊢ F Fn A ∧ S ∈ P ∧ X ∈ S ∧ y ∈ S ∧ S = F -1 F ⁡ y → F F -1 F ⁡ y = F ⁡ y ∩ ran ⁡ F
17 1 elsetpreimafvbi ⊢ F Fn A ∧ S ∈ P ∧ X ∈ S → y ∈ S ↔ y ∈ A ∧ F ⁡ y = F ⁡ X
18 fnfvelrn ⊢ F Fn A ∧ y ∈ A → F ⁡ y ∈ ran ⁡ F
19 18 snssd ⊢ F Fn A ∧ y ∈ A → F ⁡ y ⊆ ran ⁡ F
20 dfss2 ⊢ F ⁡ y ⊆ ran ⁡ F ↔ F ⁡ y ∩ ran ⁡ F = F ⁡ y
21 19 20 sylib ⊢ F Fn A ∧ y ∈ A → F ⁡ y ∩ ran ⁡ F = F ⁡ y
22 21 3adant3 ⊢ F Fn A ∧ y ∈ A ∧ F ⁡ y = F ⁡ X → F ⁡ y ∩ ran ⁡ F = F ⁡ y
23 simp3 ⊢ F Fn A ∧ y ∈ A ∧ F ⁡ y = F ⁡ X → F ⁡ y = F ⁡ X
24 23 sneqd ⊢ F Fn A ∧ y ∈ A ∧ F ⁡ y = F ⁡ X → F ⁡ y = F ⁡ X
25 22 24 eqtrd ⊢ F Fn A ∧ y ∈ A ∧ F ⁡ y = F ⁡ X → F ⁡ y ∩ ran ⁡ F = F ⁡ X
26 25 3expib ⊢ F Fn A → y ∈ A ∧ F ⁡ y = F ⁡ X → F ⁡ y ∩ ran ⁡ F = F ⁡ X
27 26 3ad2ant1 ⊢ F Fn A ∧ S ∈ P ∧ X ∈ S → y ∈ A ∧ F ⁡ y = F ⁡ X → F ⁡ y ∩ ran ⁡ F = F ⁡ X
28 17 27 sylbid ⊢ F Fn A ∧ S ∈ P ∧ X ∈ S → y ∈ S → F ⁡ y ∩ ran ⁡ F = F ⁡ X
29 28 imp ⊢ F Fn A ∧ S ∈ P ∧ X ∈ S ∧ y ∈ S → F ⁡ y ∩ ran ⁡ F = F ⁡ X
30 29 3adant3 ⊢ F Fn A ∧ S ∈ P ∧ X ∈ S ∧ y ∈ S ∧ S = F -1 F ⁡ y → F ⁡ y ∩ ran ⁡ F = F ⁡ X
31 11 16 30 3eqtrd ⊢ F Fn A ∧ S ∈ P ∧ X ∈ S ∧ y ∈ S ∧ S = F -1 F ⁡ y → F S = F ⁡ X
32 31 rexlimdv3a ⊢ F Fn A ∧ S ∈ P ∧ X ∈ S → ∃ y ∈ S S = F -1 F ⁡ y → F S = F ⁡ X
33 9 32 mpd ⊢ F Fn A ∧ S ∈ P ∧ X ∈ S → F S = F ⁡ X