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Theorem ssprss 4588
Description: A pair as subset of a pair. (Contributed by AV, 26-Oct-2020.)
Assertion
Ref Expression
ssprss ((𝐴𝑉𝐵𝑊) → ({𝐴, 𝐵} ⊆ {𝐶, 𝐷} ↔ ((𝐴 = 𝐶𝐴 = 𝐷) ∧ (𝐵 = 𝐶𝐵 = 𝐷))))

Proof of Theorem ssprss
StepHypRef Expression
1 prssg 4583 . 2 ((𝐴𝑉𝐵𝑊) → ((𝐴 ∈ {𝐶, 𝐷} ∧ 𝐵 ∈ {𝐶, 𝐷}) ↔ {𝐴, 𝐵} ⊆ {𝐶, 𝐷}))
2 elprg 4419 . . 3 (𝐴𝑉 → (𝐴 ∈ {𝐶, 𝐷} ↔ (𝐴 = 𝐶𝐴 = 𝐷)))
3 elprg 4419 . . 3 (𝐵𝑊 → (𝐵 ∈ {𝐶, 𝐷} ↔ (𝐵 = 𝐶𝐵 = 𝐷)))
42, 3bi2anan9 629 . 2 ((𝐴𝑉𝐵𝑊) → ((𝐴 ∈ {𝐶, 𝐷} ∧ 𝐵 ∈ {𝐶, 𝐷}) ↔ ((𝐴 = 𝐶𝐴 = 𝐷) ∧ (𝐵 = 𝐶𝐵 = 𝐷))))
51, 4bitr3d 273 1 ((𝐴𝑉𝐵𝑊) → ({𝐴, 𝐵} ⊆ {𝐶, 𝐷} ↔ ((𝐴 = 𝐶𝐴 = 𝐷) ∧ (𝐵 = 𝐶𝐵 = 𝐷))))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 198  wa 386  wo 836   = wceq 1601  wcel 2107  wss 3792  {cpr 4400
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1839  ax-4 1853  ax-5 1953  ax-6 2021  ax-7 2055  ax-9 2116  ax-10 2135  ax-11 2150  ax-12 2163  ax-ext 2754
This theorem depends on definitions:  df-bi 199  df-an 387  df-or 837  df-3an 1073  df-tru 1605  df-ex 1824  df-nf 1828  df-sb 2012  df-clab 2764  df-cleq 2770  df-clel 2774  df-nfc 2921  df-v 3400  df-un 3797  df-in 3799  df-ss 3806  df-sn 4399  df-pr 4401
This theorem is referenced by:  ssprsseq  4589
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