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Theorem eupth2lem1 16308
Description: Lemma for eupth2 . (Contributed by Mario Carneiro, 8-Apr-2015.)
Assertion
Ref Expression
eupth2lem1 (𝑈𝑉 → (𝑈 ∈ if(𝐴 = 𝐵, ∅, {𝐴, 𝐵}) ↔ (𝐴𝐵 ∧ (𝑈 = 𝐴𝑈 = 𝐵))))

Proof of Theorem eupth2lem1
StepHypRef Expression
1 elif 3617 . . 3 (𝑈 ∈ if(𝐴 = 𝐵, ∅, {𝐴, 𝐵}) ↔ ((𝐴 = 𝐵𝑈 ∈ ∅) ∨ (¬ 𝐴 = 𝐵𝑈 ∈ {𝐴, 𝐵})))
2 noel 3498 . . . . 5 ¬ 𝑈 ∈ ∅
32intnan 936 . . . 4 ¬ (𝐴 = 𝐵𝑈 ∈ ∅)
4 biorf 751 . . . 4 (¬ (𝐴 = 𝐵𝑈 ∈ ∅) → ((¬ 𝐴 = 𝐵𝑈 ∈ {𝐴, 𝐵}) ↔ ((𝐴 = 𝐵𝑈 ∈ ∅) ∨ (¬ 𝐴 = 𝐵𝑈 ∈ {𝐴, 𝐵}))))
53, 4ax-mp 5 . . 3 ((¬ 𝐴 = 𝐵𝑈 ∈ {𝐴, 𝐵}) ↔ ((𝐴 = 𝐵𝑈 ∈ ∅) ∨ (¬ 𝐴 = 𝐵𝑈 ∈ {𝐴, 𝐵})))
61, 5bitr4i 187 . 2 (𝑈 ∈ if(𝐴 = 𝐵, ∅, {𝐴, 𝐵}) ↔ (¬ 𝐴 = 𝐵𝑈 ∈ {𝐴, 𝐵}))
7 df-ne 2403 . . . . 5 (𝐴𝐵 ↔ ¬ 𝐴 = 𝐵)
87bicomi 132 . . . 4 𝐴 = 𝐵𝐴𝐵)
98a1i 9 . . 3 (𝑈𝑉 → (¬ 𝐴 = 𝐵𝐴𝐵))
10 elprg 3689 . . 3 (𝑈𝑉 → (𝑈 ∈ {𝐴, 𝐵} ↔ (𝑈 = 𝐴𝑈 = 𝐵)))
119, 10anbi12d 473 . 2 (𝑈𝑉 → ((¬ 𝐴 = 𝐵𝑈 ∈ {𝐴, 𝐵}) ↔ (𝐴𝐵 ∧ (𝑈 = 𝐴𝑈 = 𝐵))))
126, 11bitrid 192 1 (𝑈𝑉 → (𝑈 ∈ if(𝐴 = 𝐵, ∅, {𝐴, 𝐵}) ↔ (𝐴𝐵 ∧ (𝑈 = 𝐴𝑈 = 𝐵))))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 104  wb 105  wo 715   = wceq 1397  wcel 2202  wne 2402  c0 3494  ifcif 3605  {cpr 3670
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-in1 619  ax-in2 620  ax-io 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213
This theorem depends on definitions:  df-bi 117  df-tru 1400  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-ne 2403  df-v 2804  df-dif 3202  df-un 3204  df-nul 3495  df-if 3606  df-sn 3675  df-pr 3676
This theorem is referenced by:  eupth2lem2dc  16309
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