Metamath Proof Explorer

Theorem lcfl5a

Description: Property of a functional with a closed kernel. TODO: Make lcfl5 etc. obsolete and rewrite without C hypothesis? (Contributed by NM, 29-Jan-2015)

Ref Expression
Hypotheses lcfl5a.h ⊢ H = LHyp ⁡ K
lcfl5a.i ⊢ I = DIsoH ⁡ K ⁡ W
lcfl5a.o ⊢ ⊥ ˙ = ocH ⁡ K ⁡ W
lcfl5a.u ⊢ U = DVecH ⁡ K ⁡ W
lcfl5a.f ⊢ F = LFnl ⁡ U
lcfl5a.l ⊢ L = LKer ⁡ U
lcfl5a.k ⊢ φ → K ∈ HL ∧ W ∈ H
lcfl5a.g ⊢ φ → G ∈ F
Assertion lcfl5a ⊢ φ → ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ G = L ⁡ G ↔ L ⁡ G ∈ ran ⁡ I

Proof

Step Hyp Ref Expression
1 lcfl5a.h ⊢ H = LHyp ⁡ K
2 lcfl5a.i ⊢ I = DIsoH ⁡ K ⁡ W
3 lcfl5a.o ⊢ ⊥ ˙ = ocH ⁡ K ⁡ W
4 lcfl5a.u ⊢ U = DVecH ⁡ K ⁡ W
5 lcfl5a.f ⊢ F = LFnl ⁡ U
6 lcfl5a.l ⊢ L = LKer ⁡ U
7 lcfl5a.k ⊢ φ → K ∈ HL ∧ W ∈ H
8 lcfl5a.g ⊢ φ → G ∈ F
9 eqid ⊢ f ∈ F | ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ f = L ⁡ f = f ∈ F | ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ f = L ⁡ f
10 9 8 lcfl1 ⊢ φ → G ∈ f ∈ F | ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ f = L ⁡ f ↔ ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ G = L ⁡ G
11 1 2 3 4 5 6 9 7 8 lcfl5 ⊢ φ → G ∈ f ∈ F | ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ f = L ⁡ f ↔ L ⁡ G ∈ ran ⁡ I
12 10 11 bitr3d ⊢ φ → ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ G = L ⁡ G ↔ L ⁡ G ∈ ran ⁡ I