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 ) )