Metamath Proof Explorer


Theorem sn-subf

Description: subf without ax-mulcom . (Contributed by SN, 5-May-2024)

Ref Expression
Assertion sn-subf − : ( ℂ × ℂ ) ⟶ ℂ

Proof

Step Hyp Ref Expression
1 subval ⊢ ( ( 𝑥 ∈ ℂ ∧ 𝑦 ∈ ℂ ) → ( 𝑥 − 𝑦 ) = ( ℩ 𝑧 ∈ ℂ ( 𝑦 + 𝑧 ) = 𝑥 ) )
2 sn-subcl ⊢ ( ( 𝑥 ∈ ℂ ∧ 𝑦 ∈ ℂ ) → ( 𝑥 − 𝑦 ) ∈ ℂ )
3 1 2 eqeltrrd ⊢ ( ( 𝑥 ∈ ℂ ∧ 𝑦 ∈ ℂ ) → ( ℩ 𝑧 ∈ ℂ ( 𝑦 + 𝑧 ) = 𝑥 ) ∈ ℂ )
4 3 rgen2 ⊢ ∀ 𝑥 ∈ ℂ ∀ 𝑦 ∈ ℂ ( ℩ 𝑧 ∈ ℂ ( 𝑦 + 𝑧 ) = 𝑥 ) ∈ ℂ
5 df-sub ⊢ − = ( 𝑥 ∈ ℂ , 𝑦 ∈ ℂ ↦ ( ℩ 𝑧 ∈ ℂ ( 𝑦 + 𝑧 ) = 𝑥 ) )
6 5 fmpo ⊢ ( ∀ 𝑥 ∈ ℂ ∀ 𝑦 ∈ ℂ ( ℩ 𝑧 ∈ ℂ ( 𝑦 + 𝑧 ) = 𝑥 ) ∈ ℂ ↔ − : ( ℂ × ℂ ) ⟶ ℂ )
7 4 6 mpbi ⊢ − : ( ℂ × ℂ ) ⟶ ℂ