Theorem preqsnd 4392

Description: Equivalence for a pair equal to a singleton, deduction form. (Contributed by Thierry Arnoux, 27-Dec-2016.)

Hypotheses

Ref	Expression
preqsnd.1	⊢ (𝜑 → 𝐴 ∈ V)
preqsnd.2	⊢ (𝜑 → 𝐵 ∈ V)
preqsnd.3	⊢ (𝜑 → 𝐶 ∈ V)

Assertion

Ref	Expression
preqsnd	⊢ (𝜑 → ({𝐴, 𝐵} = {𝐶} ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐶)))

Proof of Theorem preqsnd

Step	Hyp	Ref	Expression
1		preqsnd.1	. 2 ⊢ (𝜑 → 𝐴 ∈ V)
2		preqsnd.2	. 2 ⊢ (𝜑 → 𝐵 ∈ V)
3		preqsnd.3	. 2 ⊢ (𝜑 → 𝐶 ∈ V)
4		dfsn2 4190	. . . 4 ⊢ {𝐶} = {𝐶, 𝐶}
5	4	eqeq2i 2634	. . 3 ⊢ ({𝐴, 𝐵} = {𝐶} ↔ {𝐴, 𝐵} = {𝐶, 𝐶})
6		preq12bg 4386	. . . 4 ⊢ (((𝐴 ∈ V ∧ 𝐵 ∈ V) ∧ (𝐶 ∈ V ∧ 𝐶 ∈ V)) → ({𝐴, 𝐵} = {𝐶, 𝐶} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐶) ∨ (𝐴 = 𝐶 ∧ 𝐵 = 𝐶))))
7		oridm 536	. . . 4 ⊢ (((𝐴 = 𝐶 ∧ 𝐵 = 𝐶) ∨ (𝐴 = 𝐶 ∧ 𝐵 = 𝐶)) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐶))
8	6, 7	syl6bb 276	. . 3 ⊢ (((𝐴 ∈ V ∧ 𝐵 ∈ V) ∧ (𝐶 ∈ V ∧ 𝐶 ∈ V)) → ({𝐴, 𝐵} = {𝐶, 𝐶} ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐶)))
9	5, 8	syl5bb 272	. 2 ⊢ (((𝐴 ∈ V ∧ 𝐵 ∈ V) ∧ (𝐶 ∈ V ∧ 𝐶 ∈ V)) → ({𝐴, 𝐵} = {𝐶} ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐶)))
10	1, 2, 3, 3, 9	syl22anc 1327	1 ⊢ (𝜑 → ({𝐴, 𝐵} = {𝐶} ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐶)))

Syntax hints:   → wi 4   ↔ wb 196   ∨ wo 383   ∧ wa 384   = wceq 1483   ∈ wcel 1990  Vcvv 3200  {csn 4177  {cpr 4179

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202  df-un 3579  df-sn 4178  df-pr 4180

This theorem is referenced by:  1loopgrnb0  26398  disjdifprg  29388