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Metamath Proof Explorer

Theorem latj12

Description: Swap 1st and 2nd members of lattice join. ( chj12 analog.) (Contributed by NM, 4-Jun-2012)

Ref Expression
Hypotheses latjass.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
latjass.j ⊢ ∨ = ( join ‘ 𝐾 )
Assertion latj12 ( ( 𝐾 ∈ Lat ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑋 ∨ ( 𝑌 ∨ 𝑍 ) ) = ( 𝑌 ∨ ( 𝑋 ∨ 𝑍 ) ) )

Proof

Step Hyp Ref Expression
1 latjass.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 latjass.j ⊢ ∨ = ( join ‘ 𝐾 )
3 1 2 latjcom ⊢ ( ( 𝐾 ∈ Lat ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 ∨ 𝑌 ) = ( 𝑌 ∨ 𝑋 ) )
4 3 3adant3r3 ⊢ ( ( 𝐾 ∈ Lat ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑋 ∨ 𝑌 ) = ( 𝑌 ∨ 𝑋 ) )
5 4 oveq1d ⊢ ( ( 𝐾 ∈ Lat ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 ∨ 𝑌 ) ∨ 𝑍 ) = ( ( 𝑌 ∨ 𝑋 ) ∨ 𝑍 ) )
6 1 2 latjass ⊢ ( ( 𝐾 ∈ Lat ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑋 ∨ 𝑌 ) ∨ 𝑍 ) = ( 𝑋 ∨ ( 𝑌 ∨ 𝑍 ) ) )
7 simpl ⊢ ( ( 𝐾 ∈ Lat ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → 𝐾 ∈ Lat )
8 simpr2 ⊢ ( ( 𝐾 ∈ Lat ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → 𝑌 ∈ 𝐵 )
9 simpr1 ⊢ ( ( 𝐾 ∈ Lat ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → 𝑋 ∈ 𝐵 )
10 simpr3 ⊢ ( ( 𝐾 ∈ Lat ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → 𝑍 ∈ 𝐵 )
11 1 2 latjass ⊢ ( ( 𝐾 ∈ Lat ∧ ( 𝑌 ∈ 𝐵 ∧ 𝑋 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑌 ∨ 𝑋 ) ∨ 𝑍 ) = ( 𝑌 ∨ ( 𝑋 ∨ 𝑍 ) ) )
12 7 8 9 10 11 syl13anc ⊢ ( ( 𝐾 ∈ Lat ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( 𝑌 ∨ 𝑋 ) ∨ 𝑍 ) = ( 𝑌 ∨ ( 𝑋 ∨ 𝑍 ) ) )
13 5 6 12 3eqtr3d ⊢ ( ( 𝐾 ∈ Lat ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑋 ∨ ( 𝑌 ∨ 𝑍 ) ) = ( 𝑌 ∨ ( 𝑋 ∨ 𝑍 ) ) )