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Theorem 2ralunsn 3880
 Description: Double restricted quantification over the union of a set and a singleton, using implicit substitution. (Contributed by Paul Chapman, 17-Nov-2012.)
Hypotheses
Ref Expression
2ralunsn.1 (x = B → (φχ))
2ralunsn.2 (y = B → (φψ))
2ralunsn.3 (x = B → (ψθ))
Assertion
Ref Expression
2ralunsn (B C → (x (A ∪ {B})y (A ∪ {B})φ ↔ ((x A y A φ x A ψ) (y A χ θ))))
Distinct variable groups:   x,A   x,B,y   x,C   χ,x   ψ,y   θ,x
Allowed substitution hints:   φ(x,y)   ψ(x)   χ(y)   θ(y)   A(y)   C(y)

Proof of Theorem 2ralunsn
StepHypRef Expression
1 2ralunsn.2 . . . 4 (y = B → (φψ))
21ralunsn 3879 . . 3 (B C → (y (A ∪ {B})φ ↔ (y A φ ψ)))
32ralbidv 2634 . 2 (B C → (x (A ∪ {B})y (A ∪ {B})φx (A ∪ {B})(y A φ ψ)))
4 2ralunsn.1 . . . . . 6 (x = B → (φχ))
54ralbidv 2634 . . . . 5 (x = B → (y A φy A χ))
6 2ralunsn.3 . . . . 5 (x = B → (ψθ))
75, 6anbi12d 691 . . . 4 (x = B → ((y A φ ψ) ↔ (y A χ θ)))
87ralunsn 3879 . . 3 (B C → (x (A ∪ {B})(y A φ ψ) ↔ (x A (y A φ ψ) (y A χ θ))))
9 r19.26 2746 . . . 4 (x A (y A φ ψ) ↔ (x A y A φ x A ψ))
109anbi1i 676 . . 3 ((x A (y A φ ψ) (y A χ θ)) ↔ ((x A y A φ x A ψ) (y A χ θ)))
118, 10syl6bb 252 . 2 (B C → (x (A ∪ {B})(y A φ ψ) ↔ ((x A y A φ x A ψ) (y A χ θ))))
123, 11bitrd 244 1 (B C → (x (A ∪ {B})y (A ∪ {B})φ ↔ ((x A y A φ x A ψ) (y A χ θ))))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358   = wceq 1642   ∈ wcel 1710  ∀wral 2614   ∪ cun 3207  {csn 3737 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 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 This theorem 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 2478  df-ral 2619  df-v 2861  df-sbc 3047  df-nin 3211  df-compl 3212  df-un 3214  df-sn 3741 This theorem is referenced by: (None)
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