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Theorem elabreximd 32762
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 2769 . . . . . 6 (𝑦 = 𝐴 → (𝑦 = 𝐵𝐴 = 𝐵))
32rexbidv 3189 . . . . 5 (𝑦 = 𝐴 → (∃𝑥𝐶 𝑦 = 𝐵 ↔ ∃𝑥𝐶 𝐴 = 𝐵))
43elabg 3638 . . . 4 (𝐴𝑉 → (𝐴 ∈ {𝑦 ∣ ∃𝑥𝐶 𝑦 = 𝐵} ↔ ∃𝑥𝐶 𝐴 = 𝐵))
51, 4syl 18 . . 3 (𝜑 → (𝐴 ∈ {𝑦 ∣ ∃𝑥𝐶 𝑦 = 𝐵} ↔ ∃𝑥𝐶 𝐴 = 𝐵))
65biimpa 481 . 2 ((𝜑𝐴 ∈ {𝑦 ∣ ∃𝑥𝐶 𝑦 = 𝐵}) → ∃𝑥𝐶 𝐴 = 𝐵)
7 elabreximd.1 . . . 4 𝑥𝜑
8 elabreximd.2 . . . 4 𝑥𝜒
9 simpr 489 . . . . . 6 (((𝜑𝑥𝐶) ∧ 𝐴 = 𝐵) → 𝐴 = 𝐵)
10 elabreximd.5 . . . . . . 7 ((𝜑𝑥𝐶) → 𝜓)
1110adantr 485 . . . . . 6 (((𝜑𝑥𝐶) ∧ 𝐴 = 𝐵) → 𝜓)
12 elabreximd.3 . . . . . . 7 (𝐴 = 𝐵 → (𝜒𝜓))
1312biimpar 482 . . . . . 6 ((𝐴 = 𝐵𝜓) → 𝜒)
149, 11, 13syl2anc 595 . . . . 5 (((𝜑𝑥𝐶) ∧ 𝐴 = 𝐵) → 𝜒)
1514exp31 424 . . . 4 (𝜑 → (𝑥𝐶 → (𝐴 = 𝐵𝜒)))
167, 8, 15rexlimd 3272 . . 3 (𝜑 → (∃𝑥𝐶 𝐴 = 𝐵𝜒))
1716imp 411 . 2 ((𝜑 ∧ ∃𝑥𝐶 𝐴 = 𝐵) → 𝜒)
186, 17syldan 602 1 ((𝜑𝐴 ∈ {𝑦 ∣ ∃𝑥𝐶 𝑦 = 𝐵}) → 𝜒)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  wa 400   = wceq 1563  wnf 1806  wcel 2145  {cab 2743  wrex 3089
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-12 2215  ax-ext 2737
This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1566  df-ex 1803  df-nf 1807  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-ral 3080  df-rex 3090
This theorem is referenced by:  elabreximdv  32763  abrexss  32764  iinabrex  32820  disjabrex  32833  disjabrexf  32834
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