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Metamath Proof Explorer

Theorem meetval2lem

Description: Lemma for meetval2 and meeteu . (Contributed by NM, 12-Sep-2018) TODO: combine this through meeteu into meetlem ?

		Ref	Expression
	Hypotheses	meetval2.b	⊢ 𝐵 = ( Base ‘ 𝐾 )
		meetval2.l	⊢ ≤ = ( le ‘ 𝐾 )
		meetval2.m	⊢ ∧ = ( meet ‘ 𝐾 )
		meetval2.k	⊢ ( 𝜑 → 𝐾 ∈ 𝑉 )
		meetval2.x	⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
		meetval2.y	⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
	Assertion	meetval2lem	⊢ ( ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( ∀ 𝑦 ∈ { 𝑋 , 𝑌 } 𝑥 ≤ 𝑦 ∧ ∀ 𝑧 ∈ 𝐵 ( ∀ 𝑦 ∈ { 𝑋 , 𝑌 } 𝑧 ≤ 𝑦 → 𝑧 ≤ 𝑥 ) ) ↔ ( ( 𝑥 ≤ 𝑋 ∧ 𝑥 ≤ 𝑌 ) ∧ ∀ 𝑧 ∈ 𝐵 ( ( 𝑧 ≤ 𝑋 ∧ 𝑧 ≤ 𝑌 ) → 𝑧 ≤ 𝑥 ) ) ) )

Proof

Step	Hyp	Ref	Expression
1		meetval2.b	⊢ 𝐵 = ( Base ‘ 𝐾 )
2		meetval2.l	⊢ ≤ = ( le ‘ 𝐾 )
3		meetval2.m	⊢ ∧ = ( meet ‘ 𝐾 )
4		meetval2.k	⊢ ( 𝜑 → 𝐾 ∈ 𝑉 )
5		meetval2.x	⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
6		meetval2.y	⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
7		breq2	⊢ ( 𝑦 = 𝑋 → ( 𝑥 ≤ 𝑦 ↔ 𝑥 ≤ 𝑋 ) )
8		breq2	⊢ ( 𝑦 = 𝑌 → ( 𝑥 ≤ 𝑦 ↔ 𝑥 ≤ 𝑌 ) )
9	7 8	ralprg	⊢ ( ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ∀ 𝑦 ∈ { 𝑋 , 𝑌 } 𝑥 ≤ 𝑦 ↔ ( 𝑥 ≤ 𝑋 ∧ 𝑥 ≤ 𝑌 ) ) )
10		breq2	⊢ ( 𝑦 = 𝑋 → ( 𝑧 ≤ 𝑦 ↔ 𝑧 ≤ 𝑋 ) )
11		breq2	⊢ ( 𝑦 = 𝑌 → ( 𝑧 ≤ 𝑦 ↔ 𝑧 ≤ 𝑌 ) )
12	10 11	ralprg	⊢ ( ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ∀ 𝑦 ∈ { 𝑋 , 𝑌 } 𝑧 ≤ 𝑦 ↔ ( 𝑧 ≤ 𝑋 ∧ 𝑧 ≤ 𝑌 ) ) )
13	12	imbi1d	⊢ ( ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( ∀ 𝑦 ∈ { 𝑋 , 𝑌 } 𝑧 ≤ 𝑦 → 𝑧 ≤ 𝑥 ) ↔ ( ( 𝑧 ≤ 𝑋 ∧ 𝑧 ≤ 𝑌 ) → 𝑧 ≤ 𝑥 ) ) )
14	13	ralbidv	⊢ ( ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ∀ 𝑧 ∈ 𝐵 ( ∀ 𝑦 ∈ { 𝑋 , 𝑌 } 𝑧 ≤ 𝑦 → 𝑧 ≤ 𝑥 ) ↔ ∀ 𝑧 ∈ 𝐵 ( ( 𝑧 ≤ 𝑋 ∧ 𝑧 ≤ 𝑌 ) → 𝑧 ≤ 𝑥 ) ) )
15	9 14	anbi12d	⊢ ( ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( ∀ 𝑦 ∈ { 𝑋 , 𝑌 } 𝑥 ≤ 𝑦 ∧ ∀ 𝑧 ∈ 𝐵 ( ∀ 𝑦 ∈ { 𝑋 , 𝑌 } 𝑧 ≤ 𝑦 → 𝑧 ≤ 𝑥 ) ) ↔ ( ( 𝑥 ≤ 𝑋 ∧ 𝑥 ≤ 𝑌 ) ∧ ∀ 𝑧 ∈ 𝐵 ( ( 𝑧 ≤ 𝑋 ∧ 𝑧 ≤ 𝑌 ) → 𝑧 ≤ 𝑥 ) ) ) )