Theorem opkelsikg 4265

Description: Membership in Kuratowski singleton image. (Contributed by SF, 13-Jan-2015.)

Assertion

Ref	Expression
opkelsikg	⊢ ((A ∈ V ∧ B ∈ W) → (⟪A, B⟫ ∈ SIk C ↔ ∃x∃y(A = {x} ∧ B = {y} ∧ ⟪x, y⟫ ∈ C)))

Distinct variable groups:   x,A,y   x,B,y   x,C,y

Allowed substitution hints:   V(x, y)   W(x, y)

Proof of Theorem opkelsikg

Dummy variables z t u are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-sik 4193	. 2 ⊢ SIk C = {z ∣ ∃t∃u(z = ⟪t, u⟫ ∧ ∃x∃y(t = {x} ∧ u = {y} ∧ ⟪x, y⟫ ∈ C))}
2		eqeq1 2359	. . . 4 ⊢ (t = A → (t = {x} ↔ A = {x}))
3	2	3anbi1d 1256	. . 3 ⊢ (t = A → ((t = {x} ∧ u = {y} ∧ ⟪x, y⟫ ∈ C) ↔ (A = {x} ∧ u = {y} ∧ ⟪x, y⟫ ∈ C)))
4	3	2exbidv 1628	. 2 ⊢ (t = A → (∃x∃y(t = {x} ∧ u = {y} ∧ ⟪x, y⟫ ∈ C) ↔ ∃x∃y(A = {x} ∧ u = {y} ∧ ⟪x, y⟫ ∈ C)))
5		eqeq1 2359	. . . 4 ⊢ (u = B → (u = {y} ↔ B = {y}))
6	5	3anbi2d 1257	. . 3 ⊢ (u = B → ((A = {x} ∧ u = {y} ∧ ⟪x, y⟫ ∈ C) ↔ (A = {x} ∧ B = {y} ∧ ⟪x, y⟫ ∈ C)))
7	6	2exbidv 1628	. 2 ⊢ (u = B → (∃x∃y(A = {x} ∧ u = {y} ∧ ⟪x, y⟫ ∈ C) ↔ ∃x∃y(A = {x} ∧ B = {y} ∧ ⟪x, y⟫ ∈ C)))
8	1, 4, 7	opkelopkabg 4246	1 ⊢ ((A ∈ V ∧ B ∈ W) → (⟪A, B⟫ ∈ SIk C ↔ ∃x∃y(A = {x} ∧ B = {y} ∧ ⟪x, y⟫ ∈ C)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358   ∧ w3a 934  ∃wex 1541   = wceq 1642   ∈ wcel 1710  {csn 3738  ⟪copk 4058   SI_k csik 4182

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334  ax-nin 4079  ax-sn 4088

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-3an 936  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-ne 2519  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-un 3215  df-dif 3216  df-ss 3260  df-nul 3552  df-sn 3742  df-pr 3743  df-opk 4059  df-sik 4193

This theorem is used by:  opksnelsik  4266  dfpw12  4302  setconslem1  4732  dfsi2  4752