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Theorem preqsndOLD 4610
 Description: Obsolete version of preqsnd 4609 as of 12-Jun-2022: Hypothesis preqsndOLD.3 is not needed. (Contributed by Thierry Arnoux, 27-Dec-2016.) (Proof modification is discouraged.) (New usage is discouraged.)
Hypotheses
Ref Expression
preqsnd.1 (𝜑𝐴 ∈ V)
preqsnd.2 (𝜑𝐵 ∈ V)
preqsndOLD.3 (𝜑𝐶 ∈ V)
Assertion
Ref Expression
preqsndOLD (𝜑 → ({𝐴, 𝐵} = {𝐶} ↔ (𝐴 = 𝐶𝐵 = 𝐶)))

Proof of Theorem preqsndOLD
StepHypRef Expression
1 preqsnd.1 . 2 (𝜑𝐴 ∈ V)
2 preqsnd.2 . 2 (𝜑𝐵 ∈ V)
3 preqsndOLD.3 . 2 (𝜑𝐶 ∈ V)
4 dfsn2 4412 . . . 4 {𝐶} = {𝐶, 𝐶}
54eqeq2i 2837 . . 3 ({𝐴, 𝐵} = {𝐶} ↔ {𝐴, 𝐵} = {𝐶, 𝐶})
6 preq12bg 4603 . . . 4 (((𝐴 ∈ V ∧ 𝐵 ∈ V) ∧ (𝐶 ∈ V ∧ 𝐶 ∈ V)) → ({𝐴, 𝐵} = {𝐶, 𝐶} ↔ ((𝐴 = 𝐶𝐵 = 𝐶) ∨ (𝐴 = 𝐶𝐵 = 𝐶))))
7 oridm 933 . . . 4 (((𝐴 = 𝐶𝐵 = 𝐶) ∨ (𝐴 = 𝐶𝐵 = 𝐶)) ↔ (𝐴 = 𝐶𝐵 = 𝐶))
86, 7syl6bb 279 . . 3 (((𝐴 ∈ V ∧ 𝐵 ∈ V) ∧ (𝐶 ∈ V ∧ 𝐶 ∈ V)) → ({𝐴, 𝐵} = {𝐶, 𝐶} ↔ (𝐴 = 𝐶𝐵 = 𝐶)))
95, 8syl5bb 275 . 2 (((𝐴 ∈ V ∧ 𝐵 ∈ V) ∧ (𝐶 ∈ V ∧ 𝐶 ∈ V)) → ({𝐴, 𝐵} = {𝐶} ↔ (𝐴 = 𝐶𝐵 = 𝐶)))
101, 2, 3, 3, 9syl22anc 872 1 (𝜑 → ({𝐴, 𝐵} = {𝐶} ↔ (𝐴 = 𝐶𝐵 = 𝐶)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 386   ∨ wo 878   = wceq 1656   ∈ wcel 2164  Vcvv 3414  {csn 4399  {cpr 4401 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1894  ax-4 1908  ax-5 2009  ax-6 2075  ax-7 2112  ax-9 2173  ax-10 2192  ax-11 2207  ax-12 2220  ax-ext 2803 This theorem depends on definitions:  df-bi 199  df-an 387  df-or 879  df-3an 1113  df-tru 1660  df-ex 1879  df-nf 1883  df-sb 2068  df-clab 2812  df-cleq 2818  df-clel 2821  df-nfc 2958  df-v 3416  df-un 3803  df-sn 4400  df-pr 4402 This theorem is referenced by: (None)
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