Metamath Proof Explorer


Theorem cplem2

Description: Lemma for the Collection Principle cp . (Contributed by NM, 17-Oct-2003) Use the Scott operation. (Revised by BTernaryTau, 19-Jul-2026)

Ref Expression
Hypothesis cplem2.1
|- A e. _V
Assertion cplem2
|- E. y A. x e. A ( B =/= (/) -> ( B i^i y ) =/= (/) )

Proof

Step Hyp Ref Expression
1 cplem2.1
 |-  A e. _V
2 scottex
 |-  Scott B e. _V
3 1 2 iunex
 |-  U_ x e. A Scott B e. _V
4 nfiu1
 |-  F/_ x U_ x e. A Scott B
5 4 nfeq2
 |-  F/ x y = U_ x e. A Scott B
6 ineq2
 |-  ( y = U_ x e. A Scott B -> ( B i^i y ) = ( B i^i U_ x e. A Scott B ) )
7 6 neeq1d
 |-  ( y = U_ x e. A Scott B -> ( ( B i^i y ) =/= (/) <-> ( B i^i U_ x e. A Scott B ) =/= (/) ) )
8 7 imbi2d
 |-  ( y = U_ x e. A Scott B -> ( ( B =/= (/) -> ( B i^i y ) =/= (/) ) <-> ( B =/= (/) -> ( B i^i U_ x e. A Scott B ) =/= (/) ) ) )
9 5 8 ralbid
 |-  ( y = U_ x e. A Scott B -> ( A. x e. A ( B =/= (/) -> ( B i^i y ) =/= (/) ) <-> A. x e. A ( B =/= (/) -> ( B i^i U_ x e. A Scott B ) =/= (/) ) ) )
10 eqid
 |-  U_ x e. A Scott B = U_ x e. A Scott B
11 10 cplem1
 |-  A. x e. A ( B =/= (/) -> ( B i^i U_ x e. A Scott B ) =/= (/) )
12 3 9 11 ceqsexv2d
 |-  E. y A. x e. A ( B =/= (/) -> ( B i^i y ) =/= (/) )