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

Theorem uneqin 3291
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 3115 . . . 4 ((𝐴𝐵) = (𝐴𝐵) → (𝐴𝐵) ⊆ (𝐴𝐵))
2 unss 3214 . . . . 5 ((𝐴 ⊆ (𝐴𝐵) ∧ 𝐵 ⊆ (𝐴𝐵)) ↔ (𝐴𝐵) ⊆ (𝐴𝐵))
3 ssin 3262 . . . . . . 7 ((𝐴𝐴𝐴𝐵) ↔ 𝐴 ⊆ (𝐴𝐵))
4 sstr 3069 . . . . . . 7 ((𝐴𝐴𝐴𝐵) → 𝐴𝐵)
53, 4sylbir 134 . . . . . 6 (𝐴 ⊆ (𝐴𝐵) → 𝐴𝐵)
6 ssin 3262 . . . . . . 7 ((𝐵𝐴𝐵𝐵) ↔ 𝐵 ⊆ (𝐴𝐵))
7 simpl 108 . . . . . . 7 ((𝐵𝐴𝐵𝐵) → 𝐵𝐴)
86, 7sylbir 134 . . . . . 6 (𝐵 ⊆ (𝐴𝐵) → 𝐵𝐴)
95, 8anim12i 334 . . . . 5 ((𝐴 ⊆ (𝐴𝐵) ∧ 𝐵 ⊆ (𝐴𝐵)) → (𝐴𝐵𝐵𝐴))
102, 9sylbir 134 . . . 4 ((𝐴𝐵) ⊆ (𝐴𝐵) → (𝐴𝐵𝐵𝐴))
111, 10syl 14 . . 3 ((𝐴𝐵) = (𝐴𝐵) → (𝐴𝐵𝐵𝐴))
12 eqss 3076 . . 3 (𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
1311, 12sylibr 133 . 2 ((𝐴𝐵) = (𝐴𝐵) → 𝐴 = 𝐵)
14 unidm 3183 . . . 4 (𝐴𝐴) = 𝐴
15 inidm 3249 . . . 4 (𝐴𝐴) = 𝐴
1614, 15eqtr4i 2136 . . 3 (𝐴𝐴) = (𝐴𝐴)
17 uneq2 3188 . . 3 (𝐴 = 𝐵 → (𝐴𝐴) = (𝐴𝐵))
18 ineq2 3235 . . 3 (𝐴 = 𝐵 → (𝐴𝐴) = (𝐴𝐵))
1916, 17, 183eqtr3a 2169 . 2 (𝐴 = 𝐵 → (𝐴𝐵) = (𝐴𝐵))
2013, 19impbii 125 1 ((𝐴𝐵) = (𝐴𝐵) ↔ 𝐴 = 𝐵)
Colors of variables: wff set class
Syntax hints:  wa 103  wb 104   = wceq 1312  cun 3033  cin 3034  wss 3035
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 681  ax-5 1404  ax-7 1405  ax-gen 1406  ax-ie1 1450  ax-ie2 1451  ax-8 1463  ax-10 1464  ax-11 1465  ax-i12 1466  ax-bndl 1467  ax-4 1468  ax-17 1487  ax-i9 1491  ax-ial 1495  ax-i5r 1496  ax-ext 2095
This theorem depends on definitions:  df-bi 116  df-tru 1315  df-nf 1418  df-sb 1717  df-clab 2100  df-cleq 2106  df-clel 2109  df-nfc 2242  df-v 2657  df-un 3039  df-in 3041  df-ss 3048
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator