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Metamath Proof Explorer

Theorem cdlemg3a

Description: Part of proof of Lemma G in Crawley p. 116, line 19. Show p \/ q = p \/ u. TODO: reformat cdleme0cp to match this, then replace with cdleme0cp . (Contributed by NM, 19-Apr-2013)

Ref Expression
Hypotheses cdlemg3.l ⊢ ≤ = ( le ‘ 𝐾 )
cdlemg3.j ⊢ ∨ = ( join ‘ 𝐾 )
cdlemg3.m ⊢ ∧ = ( meet ‘ 𝐾 )
cdlemg3.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
cdlemg3.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
cdlemg3.u ⊢ 𝑈 = ( ( 𝑃 ∨ 𝑄 ) ∧ 𝑊 )
Assertion cdlemg3a ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ∧ 𝑄 ∈ 𝐴 ) → ( 𝑃 ∨ 𝑄 ) = ( 𝑃 ∨ 𝑈 ) )

Proof

Step Hyp Ref Expression
1 cdlemg3.l ⊢ ≤ = ( le ‘ 𝐾 )
2 cdlemg3.j ⊢ ∨ = ( join ‘ 𝐾 )
3 cdlemg3.m ⊢ ∧ = ( meet ‘ 𝐾 )
4 cdlemg3.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
5 cdlemg3.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
6 cdlemg3.u ⊢ 𝑈 = ( ( 𝑃 ∨ 𝑄 ) ∧ 𝑊 )
7 1 2 3 4 5 6 cdleme8 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ∧ 𝑄 ∈ 𝐴 ) → ( 𝑃 ∨ 𝑈 ) = ( 𝑃 ∨ 𝑄 ) )
8 7 eqcomd ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ∧ 𝑄 ∈ 𝐴 ) → ( 𝑃 ∨ 𝑄 ) = ( 𝑃 ∨ 𝑈 ) )