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Metamath Proof Explorer

Theorem ltrnj

Description: Lattice translation of a meet. TODO: change antecedent to K e. HL (Contributed by NM, 25-May-2012)

Ref Expression
Hypotheses ltrnj.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
ltrnj.j ⊢ ∨ = ( join ‘ 𝐾 )
ltrnj.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
ltrnj.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
Assertion ltrnj ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( 𝐹 ‘ ( 𝑋 ∨ 𝑌 ) ) = ( ( 𝐹 ‘ 𝑋 ) ∨ ( 𝐹 ‘ 𝑌 ) ) )

Proof

Step Hyp Ref Expression
1 ltrnj.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 ltrnj.j ⊢ ∨ = ( join ‘ 𝐾 )
3 ltrnj.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
4 ltrnj.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
5 simp1l ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → 𝐾 ∈ HL )
6 5 hllatd ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → 𝐾 ∈ Lat )
7 eqid ⊢ ( LAut ‘ 𝐾 ) = ( LAut ‘ 𝐾 )
8 3 7 4 ltrnlaut ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ) → 𝐹 ∈ ( LAut ‘ 𝐾 ) )
9 8 3adant3 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → 𝐹 ∈ ( LAut ‘ 𝐾 ) )
10 simp3l ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → 𝑋 ∈ 𝐵 )
11 simp3r ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → 𝑌 ∈ 𝐵 )
12 1 2 7 lautj ⊢ ( ( 𝐾 ∈ Lat ∧ ( 𝐹 ∈ ( LAut ‘ 𝐾 ) ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( 𝐹 ‘ ( 𝑋 ∨ 𝑌 ) ) = ( ( 𝐹 ‘ 𝑋 ) ∨ ( 𝐹 ‘ 𝑌 ) ) )
13 6 9 10 11 12 syl13anc ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( 𝐹 ‘ ( 𝑋 ∨ 𝑌 ) ) = ( ( 𝐹 ‘ 𝑋 ) ∨ ( 𝐹 ‘ 𝑌 ) ) )