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Theorem 2reu5lem1 3041
 Description: Lemma for 2reu5 3044. Note that ∃!x ∈ A∃!y ∈ Bφ does not mean "there is exactly one x in A and exactly one y in B such that φ holds;" see comment for 2eu5 2288. (Contributed by Alexander van der Vekens, 17-Jun-2017.)
Assertion
Ref Expression
2reu5lem1 (∃!x A ∃!y B φ∃!x∃!y(x A y B φ))
Distinct variable groups:   y,A   x,B   x,y
Allowed substitution hints:   φ(x,y)   A(x)   B(y)

Proof of Theorem 2reu5lem1
StepHypRef Expression
1 df-reu 2621 . . 3 (∃!y B φ∃!y(y B φ))
21reubii 2797 . 2 (∃!x A ∃!y B φ∃!x A ∃!y(y B φ))
3 df-reu 2621 . . 3 (∃!x A ∃!y(y B φ) ↔ ∃!x(x A ∃!y(y B φ)))
4 euanv 2265 . . . . . 6 (∃!y(x A (y B φ)) ↔ (x A ∃!y(y B φ)))
54bicomi 193 . . . . 5 ((x A ∃!y(y B φ)) ↔ ∃!y(x A (y B φ)))
6 3anass 938 . . . . . . 7 ((x A y B φ) ↔ (x A (y B φ)))
76bicomi 193 . . . . . 6 ((x A (y B φ)) ↔ (x A y B φ))
87eubii 2213 . . . . 5 (∃!y(x A (y B φ)) ↔ ∃!y(x A y B φ))
95, 8bitri 240 . . . 4 ((x A ∃!y(y B φ)) ↔ ∃!y(x A y B φ))
109eubii 2213 . . 3 (∃!x(x A ∃!y(y B φ)) ↔ ∃!x∃!y(x A y B φ))
113, 10bitri 240 . 2 (∃!x A ∃!y(y B φ) ↔ ∃!x∃!y(x A y B φ))
122, 11bitri 240 1 (∃!x A ∃!y B φ∃!x∃!y(x A y B φ))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 176   ∧ wa 358   ∧ w3a 934   ∈ wcel 1710  ∃!weu 2204  ∃!wreu 2616 This theorem was proved from 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 This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-3an 936  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-eu 2208  df-mo 2209  df-reu 2621 This theorem is referenced by:  2reu5lem3  3043
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