Theorem eqfvelsetpreimafv 48617

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 48615	. 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 1569   ∈ wcel 2146  {cab 2741  ∃wrex 3090  {csn 4588  ^◡ccnv 5654   “ cima 5658   Fn wfn 6542  ‘cfv 6547

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-10 2179  ax-12 2215  ax-ext 2735  ax-sep 5252  ax-nul 5263  ax-pr 5395

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2101  df-mo 2567  df-eu 2597  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-uni 4871  df-br 5107  df-opab 5171  df-id 5550  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-rn 5666  df-res 5667  df-ima 5668  df-iota 6501  df-fun 6549  df-fn 6550  df-fv 6555

This theorem is used by: (None)