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Theorem elabreximd 6350
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 2245 . . . . . 6 (𝑦 = 𝐴 → (𝑦 = 𝐵𝐴 = 𝐵))
32rexbidv 2551 . . . . 5 (𝑦 = 𝐴 → (∃𝑥𝐶 𝑦 = 𝐵 ↔ ∃𝑥𝐶 𝐴 = 𝐵))
43elabg 2972 . . . 4 (𝐴𝑉 → (𝐴 ∈ {𝑦 ∣ ∃𝑥𝐶 𝑦 = 𝐵} ↔ ∃𝑥𝐶 𝐴 = 𝐵))
51, 4syl 14 . . 3 (𝜑 → (𝐴 ∈ {𝑦 ∣ ∃𝑥𝐶 𝑦 = 𝐵} ↔ ∃𝑥𝐶 𝐴 = 𝐵))
65biimpa 296 . 2 ((𝜑𝐴 ∈ {𝑦 ∣ ∃𝑥𝐶 𝑦 = 𝐵}) → ∃𝑥𝐶 𝐴 = 𝐵)
7 elabreximd.1 . . . 4 𝑥𝜑
8 elabreximd.2 . . . 4 𝑥𝜒
9 simpr 110 . . . . . 6 (((𝜑𝑥𝐶) ∧ 𝐴 = 𝐵) → 𝐴 = 𝐵)
10 elabreximd.5 . . . . . . 7 ((𝜑𝑥𝐶) → 𝜓)
1110adantr 276 . . . . . 6 (((𝜑𝑥𝐶) ∧ 𝐴 = 𝐵) → 𝜓)
12 elabreximd.3 . . . . . . 7 (𝐴 = 𝐵 → (𝜒𝜓))
1312biimpar 297 . . . . . 6 ((𝐴 = 𝐵𝜓) → 𝜒)
149, 11, 13syl2anc 415 . . . . 5 (((𝜑𝑥𝐶) ∧ 𝐴 = 𝐵) → 𝜒)
1514exp31 364 . . . 4 (𝜑 → (𝑥𝐶 → (𝐴 = 𝐵𝜒)))
167, 8, 15rexlimd 2665 . . 3 (𝜑 → (∃𝑥𝐶 𝐴 = 𝐵𝜒))
1716imp 124 . 2 ((𝜑 ∧ ∃𝑥𝐶 𝐴 = 𝐵) → 𝜒)
186, 17syldan 282 1 ((𝜑𝐴 ∈ {𝑦 ∣ ∃𝑥𝐶 𝑦 = 𝐵}) → 𝜒)
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104  wb 105   = wceq 1402  wnf 1513  wcel 2209  {cab 2224  wrex 2529
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220
This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-rex 2534  df-v 2823
This theorem is referenced by:  elabreximdv  6351  abrexss  6352
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