Theorem eqfvelsetpreimafv 48202

Description: If an element of the domain of the function has the same function value as an element of the preimage of a function value, then it is an element of the same preimage. (Contributed by AV, 9-Mar-2024.)

Hypothesis

Ref	Expression
setpreimafvex.p	⊢ 𝑃 = {𝑧 ∣ ∃𝑥 ∈ 𝐴 𝑧 = (◡𝐹 “ {(𝐹‘𝑥)})}

Assertion

Ref	Expression
eqfvelsetpreimafv	⊢ ((𝐹 Fn 𝐴 ∧ 𝑆 ∈ 𝑃 ∧ 𝑋 ∈ 𝑆) → ((𝑌 ∈ 𝐴 ∧ (𝐹‘𝑌) = (𝐹‘𝑋)) → 𝑌 ∈ 𝑆))

Distinct variable groups:   𝑥,𝐴,𝑧   𝑥,𝐹,𝑧   𝑥,𝑆,𝑧   𝑥,𝑋   𝑥,𝑌

Allowed substitution hints:   𝑃(𝑥, 𝑧)   𝑋(𝑧)   𝑌(𝑧)

Proof of Theorem eqfvelsetpreimafv

Step	Hyp	Ref	Expression
1		setpreimafvex.p	. . 3 ⊢ 𝑃 = {𝑧 ∣ ∃𝑥 ∈ 𝐴 𝑧 = (◡𝐹 “ {(𝐹‘𝑥)})}
2	1	elsetpreimafvbi 48200	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ 𝑆 ∈ 𝑃 ∧ 𝑋 ∈ 𝑆) → (𝑌 ∈ 𝑆 ↔ (𝑌 ∈ 𝐴 ∧ (𝐹‘𝑌) = (𝐹‘𝑋))))
3	2	biimprd 251	1 ⊢ ((𝐹 Fn 𝐴 ∧ 𝑆 ∈ 𝑃 ∧ 𝑋 ∈ 𝑆) → ((𝑌 ∈ 𝐴 ∧ (𝐹‘𝑌) = (𝐹‘𝑋)) → 𝑌 ∈ 𝑆))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ∧ w3a 1103   = wceq 1570   ∈ wcel 2146  {cab 2743  ∃wrex 3091  {csn 4591  ^◡ccnv 5662   “ cima 5666   Fn wfn 6535  ‘cfv 6540

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-10 2179  ax-12 2216  ax-ext 2737  ax-sep 5259  ax-nul 5271  ax-pr 5406

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-mo 2569  df-eu 2599  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-uni 4875  df-br 5112  df-opab 5176  df-id 5558  df-xp 5669  df-rel 5670  df-cnv 5671  df-co 5672  df-dm 5673  df-rn 5674  df-res 5675  df-ima 5676  df-iota 6496  df-fun 6542  df-fn 6543  df-fv 6548

This theorem is used by: (None)