Theorem preqsnd 4330

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 4138	. . . 4 ⊢ {𝐶} = {𝐶, 𝐶}
5	4	eqeq2i 2622	. . 3 ⊢ ({𝐴, 𝐵} = {𝐶} ↔ {𝐴, 𝐵} = {𝐶, 𝐶})
6		preq12bg 4326	. . . 4 ⊢ (((𝐴 ∈ V ∧ 𝐵 ∈ V) ∧ (𝐶 ∈ V ∧ 𝐶 ∈ V)) → ({𝐴, 𝐵} = {𝐶, 𝐶} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐶) ∨ (𝐴 = 𝐶 ∧ 𝐵 = 𝐶))))
7		oridm 535	. . . 4 ⊢ (((𝐴 = 𝐶 ∧ 𝐵 = 𝐶) ∨ (𝐴 = 𝐶 ∧ 𝐵 = 𝐶)) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐶))
8	6, 7	syl6bb 275	. . 3 ⊢ (((𝐴 ∈ V ∧ 𝐵 ∈ V) ∧ (𝐶 ∈ V ∧ 𝐶 ∈ V)) → ({𝐴, 𝐵} = {𝐶, 𝐶} ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐶)))
9	5, 8	syl5bb 271	. 2 ⊢ (((𝐴 ∈ V ∧ 𝐵 ∈ V) ∧ (𝐶 ∈ V ∧ 𝐶 ∈ V)) → ({𝐴, 𝐵} = {𝐶} ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐶)))
10	1, 2, 3, 3, 9	syl22anc 1319	1 ⊢ (𝜑 → ({𝐴, 𝐵} = {𝐶} ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐶)))

Syntax hints:   → wi 4   ↔ wb 195   ∨ wo 382   ∧ wa 383   = wceq 1475   ∈ wcel 1977  Vcvv 3173  {csn 4125  {cpr 4127

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-v 3175  df-un 3545  df-sn 4126  df-pr 4128

This theorem is referenced by:  disjdifprg  28770  1loopgrnb0  40717