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Metamath Proof Explorer

Theorem ac6gf

Description: Axiom of Choice. (Contributed by Jeff Madsen, 2-Sep-2009)

Ref Expression
Hypotheses ac6gf.1 ⊢ Ⅎ 𝑦 𝜓
ac6gf.2 ⊢ ( 𝑦 = ( 𝑓 ‘ 𝑥 ) → ( 𝜑 ↔ 𝜓 ) )
Assertion ac6gf ( ( 𝐴 ∈ 𝐶 ∧ ∀ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 ) → ∃ 𝑓 ( 𝑓 : 𝐴 ⟶ 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 𝜓 ) )

Proof

Step Hyp Ref Expression
1 ac6gf.1 ⊢ Ⅎ 𝑦 𝜓
2 ac6gf.2 ⊢ ( 𝑦 = ( 𝑓 ‘ 𝑥 ) → ( 𝜑 ↔ 𝜓 ) )
3 cbvrexsvw ⊢ ( ∃ 𝑦 ∈ 𝐵 𝜑 ↔ ∃ 𝑧 ∈ 𝐵 [ 𝑧 / 𝑦 ] 𝜑 )
4 3 ralbii ⊢ ( ∀ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 ↔ ∀ 𝑥 ∈ 𝐴 ∃ 𝑧 ∈ 𝐵 [ 𝑧 / 𝑦 ] 𝜑 )
5 1 2 sbhypf ⊢ ( 𝑧 = ( 𝑓 ‘ 𝑥 ) → ( [ 𝑧 / 𝑦 ] 𝜑 ↔ 𝜓 ) )
6 5 ac6sg ⊢ ( 𝐴 ∈ 𝐶 → ( ∀ 𝑥 ∈ 𝐴 ∃ 𝑧 ∈ 𝐵 [ 𝑧 / 𝑦 ] 𝜑 → ∃ 𝑓 ( 𝑓 : 𝐴 ⟶ 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 𝜓 ) ) )
7 6 imp ⊢ ( ( 𝐴 ∈ 𝐶 ∧ ∀ 𝑥 ∈ 𝐴 ∃ 𝑧 ∈ 𝐵 [ 𝑧 / 𝑦 ] 𝜑 ) → ∃ 𝑓 ( 𝑓 : 𝐴 ⟶ 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 𝜓 ) )
8 4 7 sylan2b ⊢ ( ( 𝐴 ∈ 𝐶 ∧ ∀ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 ) → ∃ 𝑓 ( 𝑓 : 𝐴 ⟶ 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 𝜓 ) )