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 ) )