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Metamath Proof Explorer

Theorem supcl

Description: A supremum belongs to its base class (closure law). See also supub and suplub . (Contributed by NM, 12-Oct-2004)

Ref Expression
Hypotheses supmo.1 ⊢ ( 𝜑 → 𝑅 Or 𝐴 )
supcl.2 ⊢ ( 𝜑 → ∃ 𝑥 ∈ 𝐴 ( ∀ 𝑦 ∈ 𝐵 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐴 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐵 𝑦 𝑅 𝑧 ) ) )
Assertion supcl ( 𝜑 → sup ( 𝐵 , 𝐴 , 𝑅 ) ∈ 𝐴 )

Proof

Step Hyp Ref Expression
1 supmo.1 ⊢ ( 𝜑 → 𝑅 Or 𝐴 )
2 supcl.2 ⊢ ( 𝜑 → ∃ 𝑥 ∈ 𝐴 ( ∀ 𝑦 ∈ 𝐵 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐴 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐵 𝑦 𝑅 𝑧 ) ) )
3 1 supval2 ⊢ ( 𝜑 → sup ( 𝐵 , 𝐴 , 𝑅 ) = ( ℩ 𝑥 ∈ 𝐴 ( ∀ 𝑦 ∈ 𝐵 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐴 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐵 𝑦 𝑅 𝑧 ) ) ) )
4 1 2 supeu ⊢ ( 𝜑 → ∃! 𝑥 ∈ 𝐴 ( ∀ 𝑦 ∈ 𝐵 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐴 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐵 𝑦 𝑅 𝑧 ) ) )
5 riotacl ⊢ ( ∃! 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 ) ) ) ∈ 𝐴 )
6 4 5 syl ⊢ ( 𝜑 → ( ℩ 𝑥 ∈ 𝐴 ( ∀ 𝑦 ∈ 𝐵 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐴 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐵 𝑦 𝑅 𝑧 ) ) ) ∈ 𝐴 )
7 3 6 eqeltrd ⊢ ( 𝜑 → sup ( 𝐵 , 𝐴 , 𝑅 ) ∈ 𝐴 )