Theorem dalem63

Description: Lemma for dath . Combine the cases where Y and Z are different planes with the case where Y and Z are the same plane. (Contributed by NM, 11-Aug-2012)

	Ref	Expression
Hypotheses	dalem62.ph	⊢ ( 𝜑 ↔ ( ( ( 𝐾 ∈ HL ∧ 𝐶 ∈ ( Base ‘ 𝐾 ) ) ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ ( 𝑆 ∈ 𝐴 ∧ 𝑇 ∈ 𝐴 ∧ 𝑈 ∈ 𝐴 ) ) ∧ ( 𝑌 ∈ 𝑂 ∧ 𝑍 ∈ 𝑂 ) ∧ ( ( ¬ 𝐶 ≤ ( 𝑃 ∨ 𝑄 ) ∧ ¬ 𝐶 ≤ ( 𝑄 ∨ 𝑅 ) ∧ ¬ 𝐶 ≤ ( 𝑅 ∨ 𝑃 ) ) ∧ ( ¬ 𝐶 ≤ ( 𝑆 ∨ 𝑇 ) ∧ ¬ 𝐶 ≤ ( 𝑇 ∨ 𝑈 ) ∧ ¬ 𝐶 ≤ ( 𝑈 ∨ 𝑆 ) ) ∧ ( 𝐶 ≤ ( 𝑃 ∨ 𝑆 ) ∧ 𝐶 ≤ ( 𝑄 ∨ 𝑇 ) ∧ 𝐶 ≤ ( 𝑅 ∨ 𝑈 ) ) ) ) )
	dalem62.l	⊢ ≤ = ( le ‘ 𝐾 )
	dalem62.j	⊢ ∨ = ( join ‘ 𝐾 )
	dalem62.a	⊢ 𝐴 = ( Atoms ‘ 𝐾 )
	dalem62.m	⊢ ∧ = ( meet ‘ 𝐾 )
	dalem62.o	⊢ 𝑂 = ( LPlanes ‘ 𝐾 )
	dalem62.y	⊢ 𝑌 = ( ( 𝑃 ∨ 𝑄 ) ∨ 𝑅 )
	dalem62.z	⊢ 𝑍 = ( ( 𝑆 ∨ 𝑇 ) ∨ 𝑈 )
	dalem62.d	⊢ 𝐷 = ( ( 𝑃 ∨ 𝑄 ) ∧ ( 𝑆 ∨ 𝑇 ) )
	dalem62.e	⊢ 𝐸 = ( ( 𝑄 ∨ 𝑅 ) ∧ ( 𝑇 ∨ 𝑈 ) )
	dalem62.f	⊢ 𝐹 = ( ( 𝑅 ∨ 𝑃 ) ∧ ( 𝑈 ∨ 𝑆 ) )
Assertion	dalem63	⊢ ( 𝜑 → 𝐹 ≤ ( 𝐷 ∨ 𝐸 ) )

Proof

Step	Hyp	Ref	Expression
1		dalem62.ph	⊢ ( 𝜑 ↔ ( ( ( 𝐾 ∈ HL ∧ 𝐶 ∈ ( Base ‘ 𝐾 ) ) ∧ ( 𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴 ) ∧ ( 𝑆 ∈ 𝐴 ∧ 𝑇 ∈ 𝐴 ∧ 𝑈 ∈ 𝐴 ) ) ∧ ( 𝑌 ∈ 𝑂 ∧ 𝑍 ∈ 𝑂 ) ∧ ( ( ¬ 𝐶 ≤ ( 𝑃 ∨ 𝑄 ) ∧ ¬ 𝐶 ≤ ( 𝑄 ∨ 𝑅 ) ∧ ¬ 𝐶 ≤ ( 𝑅 ∨ 𝑃 ) ) ∧ ( ¬ 𝐶 ≤ ( 𝑆 ∨ 𝑇 ) ∧ ¬ 𝐶 ≤ ( 𝑇 ∨ 𝑈 ) ∧ ¬ 𝐶 ≤ ( 𝑈 ∨ 𝑆 ) ) ∧ ( 𝐶 ≤ ( 𝑃 ∨ 𝑆 ) ∧ 𝐶 ≤ ( 𝑄 ∨ 𝑇 ) ∧ 𝐶 ≤ ( 𝑅 ∨ 𝑈 ) ) ) ) )
2		dalem62.l	⊢ ≤ = ( le ‘ 𝐾 )
3		dalem62.j	⊢ ∨ = ( join ‘ 𝐾 )
4		dalem62.a	⊢ 𝐴 = ( Atoms ‘ 𝐾 )
5		dalem62.m	⊢ ∧ = ( meet ‘ 𝐾 )
6		dalem62.o	⊢ 𝑂 = ( LPlanes ‘ 𝐾 )
7		dalem62.y	⊢ 𝑌 = ( ( 𝑃 ∨ 𝑄 ) ∨ 𝑅 )
8		dalem62.z	⊢ 𝑍 = ( ( 𝑆 ∨ 𝑇 ) ∨ 𝑈 )
9		dalem62.d	⊢ 𝐷 = ( ( 𝑃 ∨ 𝑄 ) ∧ ( 𝑆 ∨ 𝑇 ) )
10		dalem62.e	⊢ 𝐸 = ( ( 𝑄 ∨ 𝑅 ) ∧ ( 𝑇 ∨ 𝑈 ) )
11		dalem62.f	⊢ 𝐹 = ( ( 𝑅 ∨ 𝑃 ) ∧ ( 𝑈 ∨ 𝑆 ) )
12	1 2 3 4 5 6 7 8 9 10 11	dalem62	⊢ ( ( 𝜑 ∧ 𝑌 = 𝑍 ) → 𝐹 ≤ ( 𝐷 ∨ 𝐸 ) )
13	1 2 3 4 5 6 7 8 9 10 11	dalem16	⊢ ( ( 𝜑 ∧ 𝑌 ≠ 𝑍 ) → 𝐹 ≤ ( 𝐷 ∨ 𝐸 ) )
14	12 13	pm2.61dane	⊢ ( 𝜑 → 𝐹 ≤ ( 𝐷 ∨ 𝐸 ) )