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Metamath Proof Explorer

Theorem trljco2

Description: Trace joined with trace of composition. (Contributed by NM, 16-Jun-2013)

Ref Expression
Hypotheses trljco.j ⊢ ∨ = ( join ‘ 𝐾 )
trljco.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
trljco.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
trljco.r ⊢ 𝑅 = ( ( trL ‘ 𝐾 ) ‘ 𝑊 )
Assertion trljco2 ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → ( ( 𝑅 ‘ 𝐹 ) ∨ ( 𝑅 ‘ ( 𝐹 ∘ 𝐺 ) ) ) = ( ( 𝑅 ‘ 𝐺 ) ∨ ( 𝑅 ‘ ( 𝐹 ∘ 𝐺 ) ) ) )

Proof

Step Hyp Ref Expression
1 trljco.j ⊢ ∨ = ( join ‘ 𝐾 )
2 trljco.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
3 trljco.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
4 trljco.r ⊢ 𝑅 = ( ( trL ‘ 𝐾 ) ‘ 𝑊 )
5 simp1l ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → 𝐾 ∈ HL )
6 5 hllatd ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → 𝐾 ∈ Lat )
7 eqid ⊢ ( Base ‘ 𝐾 ) = ( Base ‘ 𝐾 )
8 7 2 3 4 trlcl ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ) → ( 𝑅 ‘ 𝐹 ) ∈ ( Base ‘ 𝐾 ) )
9 8 3adant3 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → ( 𝑅 ‘ 𝐹 ) ∈ ( Base ‘ 𝐾 ) )
10 7 2 3 4 trlcl ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐺 ∈ 𝑇 ) → ( 𝑅 ‘ 𝐺 ) ∈ ( Base ‘ 𝐾 ) )
11 10 3adant2 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → ( 𝑅 ‘ 𝐺 ) ∈ ( Base ‘ 𝐾 ) )
12 7 1 latjcom ⊢ ( ( 𝐾 ∈ Lat ∧ ( 𝑅 ‘ 𝐹 ) ∈ ( Base ‘ 𝐾 ) ∧ ( 𝑅 ‘ 𝐺 ) ∈ ( Base ‘ 𝐾 ) ) → ( ( 𝑅 ‘ 𝐹 ) ∨ ( 𝑅 ‘ 𝐺 ) ) = ( ( 𝑅 ‘ 𝐺 ) ∨ ( 𝑅 ‘ 𝐹 ) ) )
13 6 9 11 12 syl3anc ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → ( ( 𝑅 ‘ 𝐹 ) ∨ ( 𝑅 ‘ 𝐺 ) ) = ( ( 𝑅 ‘ 𝐺 ) ∨ ( 𝑅 ‘ 𝐹 ) ) )
14 1 2 3 4 trljco ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐺 ∈ 𝑇 ∧ 𝐹 ∈ 𝑇 ) → ( ( 𝑅 ‘ 𝐺 ) ∨ ( 𝑅 ‘ ( 𝐺 ∘ 𝐹 ) ) ) = ( ( 𝑅 ‘ 𝐺 ) ∨ ( 𝑅 ‘ 𝐹 ) ) )
15 14 3com23 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → ( ( 𝑅 ‘ 𝐺 ) ∨ ( 𝑅 ‘ ( 𝐺 ∘ 𝐹 ) ) ) = ( ( 𝑅 ‘ 𝐺 ) ∨ ( 𝑅 ‘ 𝐹 ) ) )
16 13 15 eqtr4d ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → ( ( 𝑅 ‘ 𝐹 ) ∨ ( 𝑅 ‘ 𝐺 ) ) = ( ( 𝑅 ‘ 𝐺 ) ∨ ( 𝑅 ‘ ( 𝐺 ∘ 𝐹 ) ) ) )
17 1 2 3 4 trljco ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → ( ( 𝑅 ‘ 𝐹 ) ∨ ( 𝑅 ‘ ( 𝐹 ∘ 𝐺 ) ) ) = ( ( 𝑅 ‘ 𝐹 ) ∨ ( 𝑅 ‘ 𝐺 ) ) )
18 2 3 ltrncom ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → ( 𝐹 ∘ 𝐺 ) = ( 𝐺 ∘ 𝐹 ) )
19 18 fveq2d ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → ( 𝑅 ‘ ( 𝐹 ∘ 𝐺 ) ) = ( 𝑅 ‘ ( 𝐺 ∘ 𝐹 ) ) )
20 19 oveq2d ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → ( ( 𝑅 ‘ 𝐺 ) ∨ ( 𝑅 ‘ ( 𝐹 ∘ 𝐺 ) ) ) = ( ( 𝑅 ‘ 𝐺 ) ∨ ( 𝑅 ‘ ( 𝐺 ∘ 𝐹 ) ) ) )
21 16 17 20 3eqtr4d ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝐹 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ) → ( ( 𝑅 ‘ 𝐹 ) ∨ ( 𝑅 ‘ ( 𝐹 ∘ 𝐺 ) ) ) = ( ( 𝑅 ‘ 𝐺 ) ∨ ( 𝑅 ‘ ( 𝐹 ∘ 𝐺 ) ) ) )