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Theorem elabreximd 30855
Description: Class substitution in an image set. (Contributed by Thierry Arnoux, 30-Dec-2016.)
Hypotheses
Ref Expression
elabreximd.1 𝑥𝜑
elabreximd.2 𝑥𝜒
elabreximd.3 (𝐴 = 𝐵 → (𝜒𝜓))
elabreximd.4 (𝜑𝐴𝑉)
elabreximd.5 ((𝜑𝑥𝐶) → 𝜓)
Assertion
Ref Expression
elabreximd ((𝜑𝐴 ∈ {𝑦 ∣ ∃𝑥𝐶 𝑦 = 𝐵}) → 𝜒)
Distinct variable groups:   𝑥,𝑦,𝐴   𝑦,𝐵   𝑦,𝐶
Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑥,𝑦)   𝜒(𝑥,𝑦)   𝐵(𝑥)   𝐶(𝑥)   𝑉(𝑥,𝑦)

Proof of Theorem elabreximd
StepHypRef Expression
1 elabreximd.4 . . . 4 (𝜑𝐴𝑉)
2 eqeq1 2742 . . . . . 6 (𝑦 = 𝐴 → (𝑦 = 𝐵𝐴 = 𝐵))
32rexbidv 3226 . . . . 5 (𝑦 = 𝐴 → (∃𝑥𝐶 𝑦 = 𝐵 ↔ ∃𝑥𝐶 𝐴 = 𝐵))
43elabg 3607 . . . 4 (𝐴𝑉 → (𝐴 ∈ {𝑦 ∣ ∃𝑥𝐶 𝑦 = 𝐵} ↔ ∃𝑥𝐶 𝐴 = 𝐵))
51, 4syl 17 . . 3 (𝜑 → (𝐴 ∈ {𝑦 ∣ ∃𝑥𝐶 𝑦 = 𝐵} ↔ ∃𝑥𝐶 𝐴 = 𝐵))
65biimpa 477 . 2 ((𝜑𝐴 ∈ {𝑦 ∣ ∃𝑥𝐶 𝑦 = 𝐵}) → ∃𝑥𝐶 𝐴 = 𝐵)
7 elabreximd.1 . . . 4 𝑥𝜑
8 elabreximd.2 . . . 4 𝑥𝜒
9 simpr 485 . . . . . 6 (((𝜑𝑥𝐶) ∧ 𝐴 = 𝐵) → 𝐴 = 𝐵)
10 elabreximd.5 . . . . . . 7 ((𝜑𝑥𝐶) → 𝜓)
1110adantr 481 . . . . . 6 (((𝜑𝑥𝐶) ∧ 𝐴 = 𝐵) → 𝜓)
12 elabreximd.3 . . . . . . 7 (𝐴 = 𝐵 → (𝜒𝜓))
1312biimpar 478 . . . . . 6 ((𝐴 = 𝐵𝜓) → 𝜒)
149, 11, 13syl2anc 584 . . . . 5 (((𝜑𝑥𝐶) ∧ 𝐴 = 𝐵) → 𝜒)
1514exp31 420 . . . 4 (𝜑 → (𝑥𝐶 → (𝐴 = 𝐵𝜒)))
167, 8, 15rexlimd 3250 . . 3 (𝜑 → (∃𝑥𝐶 𝐴 = 𝐵𝜒))
1716imp 407 . 2 ((𝜑 ∧ ∃𝑥𝐶 𝐴 = 𝐵) → 𝜒)
186, 17syldan 591 1 ((𝜑𝐴 ∈ {𝑦 ∣ ∃𝑥𝐶 𝑦 = 𝐵}) → 𝜒)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wa 396   = wceq 1539  wnf 1786  wcel 2106  {cab 2715  wrex 3065
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-12 2171  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-ex 1783  df-nf 1787  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ral 3069  df-rex 3070
This theorem is referenced by:  elabreximdv  30856  abrexss  30857  iinabrex  30908  disjabrex  30921  disjabrexf  30922
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