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Theorem 2ralunsn 3560
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 3559 . . 3 (B 𝐶 → (y (A ∪ {B})φ ↔ (y A φ ψ)))
32ralbidv 2320 . 2 (B 𝐶 → (x (A ∪ {B})y (A ∪ {B})φx (A ∪ {B})(y A φ ψ)))
4 2ralunsn.1 . . . . . 6 (x = B → (φχ))
54ralbidv 2320 . . . . 5 (x = B → (y A φy A χ))
6 2ralunsn.3 . . . . 5 (x = B → (ψθ))
75, 6anbi12d 442 . . . 4 (x = B → ((y A φ ψ) ↔ (y A χ θ)))
87ralunsn 3559 . . 3 (B 𝐶 → (x (A ∪ {B})(y A φ ψ) ↔ (x A (y A φ ψ) (y A χ θ))))
9 r19.26 2435 . . . 4 (x A (y A φ ψ) ↔ (x A y A φ x A ψ))
109anbi1i 431 . . 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 185 . 2 (B 𝐶 → (x (A ∪ {B})(y A φ ψ) ↔ ((x A y A φ x A ψ) (y A χ θ))))
123, 11bitrd 177 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 set class
Syntax hints:  wi 4   wa 97  wb 98   = wceq 1242   wcel 1390  wral 2300  cun 2909  {csn 3367
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-3an 886  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-ral 2305  df-v 2553  df-sbc 2759  df-un 2916  df-sn 3373
This theorem is referenced by: (None)
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