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 ) ) ) )