Metamath Proof Explorer


Theorem minveclem5

Description: Lemma for minvec . Discharge the assumptions in minveclem4 . (Contributed by Mario Carneiro, 9-May-2014) (Revised by Mario Carneiro, 15-Oct-2015)

Ref Expression
Hypotheses minvec.x X = Base U
minvec.m - ˙ = - U
minvec.n N = norm U
minvec.u φ U CPreHil
minvec.y φ Y LSubSp U
minvec.w φ U 𝑠 Y CMetSp
minvec.a φ A X
minvec.j J = TopOpen U
minvec.r R = ran y Y N A - ˙ y
minvec.s S = sup R <
minvec.d D = dist U X × X
Assertion minveclem5 φ x Y y Y N A - ˙ x N A - ˙ y

Proof

Step Hyp Ref Expression
1 minvec.x X = Base U
2 minvec.m - ˙ = - U
3 minvec.n N = norm U
4 minvec.u φ U CPreHil
5 minvec.y φ Y LSubSp U
6 minvec.w φ U 𝑠 Y CMetSp
7 minvec.a φ A X
8 minvec.j J = TopOpen U
9 minvec.r R = ran y Y N A - ˙ y
10 minvec.s S = sup R <
11 minvec.d D = dist U X × X
12 oveq2 s = r S 2 + s = S 2 + r
13 12 breq2d s = r A D z 2 S 2 + s A D z 2 S 2 + r
14 13 rabbidv s = r z Y | A D z 2 S 2 + s = z Y | A D z 2 S 2 + r
15 oveq2 z = y A D z = A D y
16 15 oveq1d z = y A D z 2 = A D y 2
17 16 breq1d z = y A D z 2 S 2 + r A D y 2 S 2 + r
18 17 cbvrabv z Y | A D z 2 S 2 + r = y Y | A D y 2 S 2 + r
19 14 18 eqtrdi s = r z Y | A D z 2 S 2 + s = y Y | A D y 2 S 2 + r
20 19 cbvmptv s + z Y | A D z 2 S 2 + s = r + y Y | A D y 2 S 2 + r
21 20 rneqi ran s + z Y | A D z 2 S 2 + s = ran r + y Y | A D y 2 S 2 + r
22 eqid J fLim X filGen ran s + z Y | A D z 2 S 2 + s = J fLim X filGen ran s + z Y | A D z 2 S 2 + s
23 eqid A D J fLim X filGen ran s + z Y | A D z 2 S 2 + s + S 2 2 S 2 = A D J fLim X filGen ran s + z Y | A D z 2 S 2 + s + S 2 2 S 2
24 1 2 3 4 5 6 7 8 9 10 11 21 22 23 minveclem4 φ x Y y Y N A - ˙ x N A - ˙ y