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Metamath Proof Explorer

Theorem ac6sf2

Description: Alternate version of ac6 with bound-variable hypothesis. (Contributed by NM, 2-Mar-2008) (Revised by Thierry Arnoux, 17-May-2020)

Ref Expression
Hypotheses ac6sf2.y ⊢ Ⅎ 𝑦 𝐵
ac6sf2.1 ⊢ Ⅎ 𝑦 𝜓
ac6sf2.2 ⊢ 𝐴 ∈ V
ac6sf2.3 ⊢ ( 𝑦 = ( 𝑓 ‘ 𝑥 ) → ( 𝜑 ↔ 𝜓 ) )
Assertion ac6sf2 ( ∀ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 → ∃ 𝑓 ( 𝑓 : 𝐴 ⟶ 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 𝜓 ) )

Proof

Step Hyp Ref Expression
1 ac6sf2.y ⊢ Ⅎ 𝑦 𝐵
2 ac6sf2.1 ⊢ Ⅎ 𝑦 𝜓
3 ac6sf2.2 ⊢ 𝐴 ∈ V
4 ac6sf2.3 ⊢ ( 𝑦 = ( 𝑓 ‘ 𝑥 ) → ( 𝜑 ↔ 𝜓 ) )
5 nfcv ⊢ Ⅎ 𝑧 𝐵
6 nfv ⊢ Ⅎ 𝑧 𝜑
7 nfs1v ⊢ Ⅎ 𝑦 [ 𝑧 / 𝑦 ] 𝜑
8 sbequ12 ⊢ ( 𝑦 = 𝑧 → ( 𝜑 ↔ [ 𝑧 / 𝑦 ] 𝜑 ) )
9 1 5 6 7 8 cbvrexfw ⊢ ( ∃ 𝑦 ∈ 𝐵 𝜑 ↔ ∃ 𝑧 ∈ 𝐵 [ 𝑧 / 𝑦 ] 𝜑 )
10 9 ralbii ⊢ ( ∀ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 ↔ ∀ 𝑥 ∈ 𝐴 ∃ 𝑧 ∈ 𝐵 [ 𝑧 / 𝑦 ] 𝜑 )
11 2 4 sbhypf ⊢ ( 𝑧 = ( 𝑓 ‘ 𝑥 ) → ( [ 𝑧 / 𝑦 ] 𝜑 ↔ 𝜓 ) )
12 3 11 ac6s ⊢ ( ∀ 𝑥 ∈ 𝐴 ∃ 𝑧 ∈ 𝐵 [ 𝑧 / 𝑦 ] 𝜑 → ∃ 𝑓 ( 𝑓 : 𝐴 ⟶ 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 𝜓 ) )
13 10 12 sylbi ⊢ ( ∀ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 → ∃ 𝑓 ( 𝑓 : 𝐴 ⟶ 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 𝜓 ) )