Metamath Proof Explorer


Theorem supclt

Description: Closure of supremum. (Contributed by Jeff Madsen, 2-Sep-2009)

Ref Expression
Assertion supclt ( ( 𝑅 Or 𝐴 ∧ ∃ 𝑥 ∈ 𝐴 ( ∀ 𝑦 ∈ 𝐵 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐴 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐵 𝑦 𝑅 𝑧 ) ) ) → sup ( 𝐵 , 𝐴 , 𝑅 ) ∈ 𝐴 )

Proof

Step Hyp Ref Expression
1 simpl ⊢ ( ( 𝑅 Or 𝐴 ∧ ∃ 𝑥 ∈ 𝐴 ( ∀ 𝑦 ∈ 𝐵 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐴 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐵 𝑦 𝑅 𝑧 ) ) ) → 𝑅 Or 𝐴 )
2 simpr ⊢ ( ( 𝑅 Or 𝐴 ∧ ∃ 𝑥 ∈ 𝐴 ( ∀ 𝑦 ∈ 𝐵 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐴 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐵 𝑦 𝑅 𝑧 ) ) ) → ∃ 𝑥 ∈ 𝐴 ( ∀ 𝑦 ∈ 𝐵 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐴 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐵 𝑦 𝑅 𝑧 ) ) )
3 1 2 supcl ⊢ ( ( 𝑅 Or 𝐴 ∧ ∃ 𝑥 ∈ 𝐴 ( ∀ 𝑦 ∈ 𝐵 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐴 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐵 𝑦 𝑅 𝑧 ) ) ) → sup ( 𝐵 , 𝐴 , 𝑅 ) ∈ 𝐴 )