ILE Home Intuitionistic Logic Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  ILE Home  >  Th. List  >  uneqin GIF version

Theorem uneqin 3458
Description: Equality of union and intersection implies equality of their arguments. (Contributed by NM, 16-Apr-2006.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
Assertion
Ref Expression
uneqin ((𝐴𝐵) = (𝐴𝐵) ↔ 𝐴 = 𝐵)

Proof of Theorem uneqin
StepHypRef Expression
1 eqimss 3281 . . . 4 ((𝐴𝐵) = (𝐴𝐵) → (𝐴𝐵) ⊆ (𝐴𝐵))
2 unss 3381 . . . . 5 ((𝐴 ⊆ (𝐴𝐵) ∧ 𝐵 ⊆ (𝐴𝐵)) ↔ (𝐴𝐵) ⊆ (𝐴𝐵))
3 ssin 3429 . . . . . . 7 ((𝐴𝐴𝐴𝐵) ↔ 𝐴 ⊆ (𝐴𝐵))
4 sstr 3235 . . . . . . 7 ((𝐴𝐴𝐴𝐵) → 𝐴𝐵)
53, 4sylbir 135 . . . . . 6 (𝐴 ⊆ (𝐴𝐵) → 𝐴𝐵)
6 ssin 3429 . . . . . . 7 ((𝐵𝐴𝐵𝐵) ↔ 𝐵 ⊆ (𝐴𝐵))
7 simpl 109 . . . . . . 7 ((𝐵𝐴𝐵𝐵) → 𝐵𝐴)
86, 7sylbir 135 . . . . . 6 (𝐵 ⊆ (𝐴𝐵) → 𝐵𝐴)
95, 8anim12i 338 . . . . 5 ((𝐴 ⊆ (𝐴𝐵) ∧ 𝐵 ⊆ (𝐴𝐵)) → (𝐴𝐵𝐵𝐴))
102, 9sylbir 135 . . . 4 ((𝐴𝐵) ⊆ (𝐴𝐵) → (𝐴𝐵𝐵𝐴))
111, 10syl 14 . . 3 ((𝐴𝐵) = (𝐴𝐵) → (𝐴𝐵𝐵𝐴))
12 eqss 3242 . . 3 (𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
1311, 12sylibr 134 . 2 ((𝐴𝐵) = (𝐴𝐵) → 𝐴 = 𝐵)
14 unidm 3350 . . . 4 (𝐴𝐴) = 𝐴
15 inidm 3416 . . . 4 (𝐴𝐴) = 𝐴
1614, 15eqtr4i 2255 . . 3 (𝐴𝐴) = (𝐴𝐴)
17 uneq2 3355 . . 3 (𝐴 = 𝐵 → (𝐴𝐴) = (𝐴𝐵))
18 ineq2 3402 . . 3 (𝐴 = 𝐵 → (𝐴𝐴) = (𝐴𝐵))
1916, 17, 183eqtr3a 2288 . 2 (𝐴 = 𝐵 → (𝐴𝐵) = (𝐴𝐵))
2013, 19impbii 126 1 ((𝐴𝐵) = (𝐴𝐵) ↔ 𝐴 = 𝐵)
Colors of variables: wff set class
Syntax hints:  wa 104  wb 105   = wceq 1397  cun 3198  cin 3199  wss 3200
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 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-v 2804  df-un 3204  df-in 3206  df-ss 3213
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator