Metamath Proof Explorer


Theorem dihglblem3aN

Description: Isomorphism H of a lattice glb. (Contributed by NM, 7-Apr-2014) (New usage is discouraged.)

Ref Expression
Hypotheses dihglblem.b 𝐵 = ( Base ‘ 𝐾 )
dihglblem.l = ( le ‘ 𝐾 )
dihglblem.m = ( meet ‘ 𝐾 )
dihglblem.g 𝐺 = ( glb ‘ 𝐾 )
dihglblem.h 𝐻 = ( LHyp ‘ 𝐾 )
dihglblem.t 𝑇 = { 𝑢𝐵 ∣ ∃ 𝑣𝑆 𝑢 = ( 𝑣 𝑊 ) }
dihglblem.i 𝐽 = ( ( DIsoB ‘ 𝐾 ) ‘ 𝑊 )
dihglblem.ih 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
Assertion dihglblem3aN ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑆𝐵𝑆 ≠ ∅ ) ∧ ( 𝐺𝑆 ) 𝑊 ) → ( 𝐼 ‘ ( 𝐺𝑆 ) ) = 𝑥𝑇 ( 𝐼𝑥 ) )

Proof

Step Hyp Ref Expression
1 dihglblem.b 𝐵 = ( Base ‘ 𝐾 )
2 dihglblem.l = ( le ‘ 𝐾 )
3 dihglblem.m = ( meet ‘ 𝐾 )
4 dihglblem.g 𝐺 = ( glb ‘ 𝐾 )
5 dihglblem.h 𝐻 = ( LHyp ‘ 𝐾 )
6 dihglblem.t 𝑇 = { 𝑢𝐵 ∣ ∃ 𝑣𝑆 𝑢 = ( 𝑣 𝑊 ) }
7 dihglblem.i 𝐽 = ( ( DIsoB ‘ 𝐾 ) ‘ 𝑊 )
8 dihglblem.ih 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
9 1 2 3 4 5 6 dihglblem2N ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑆𝐵 ∧ ( 𝐺𝑆 ) 𝑊 ) → ( 𝐺𝑆 ) = ( 𝐺𝑇 ) )
10 9 3adant2r ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑆𝐵𝑆 ≠ ∅ ) ∧ ( 𝐺𝑆 ) 𝑊 ) → ( 𝐺𝑆 ) = ( 𝐺𝑇 ) )
11 10 fveq2d ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑆𝐵𝑆 ≠ ∅ ) ∧ ( 𝐺𝑆 ) 𝑊 ) → ( 𝐼 ‘ ( 𝐺𝑆 ) ) = ( 𝐼 ‘ ( 𝐺𝑇 ) ) )
12 1 2 3 4 5 6 7 8 dihglblem3N ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑆𝐵𝑆 ≠ ∅ ) ∧ ( 𝐺𝑆 ) 𝑊 ) → ( 𝐼 ‘ ( 𝐺𝑇 ) ) = 𝑥𝑇 ( 𝐼𝑥 ) )
13 11 12 eqtrd ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑆𝐵𝑆 ≠ ∅ ) ∧ ( 𝐺𝑆 ) 𝑊 ) → ( 𝐼 ‘ ( 𝐺𝑆 ) ) = 𝑥𝑇 ( 𝐼𝑥 ) )