Theorem sn-negex

Description: Proof of cnegex without ax-mulcom . (Contributed by SN, 30-Apr-2024)

	Ref	Expression
Assertion	sn-negex	⊢ ( 𝐴 ∈ ℂ → ∃ 𝑏 ∈ ℂ ( 𝐴 + 𝑏 ) = 0 )

Proof

Step	Hyp	Ref	Expression
1		sn-negex12	⊢ ( 𝐴 ∈ ℂ → ∃ 𝑏 ∈ ℂ ( ( 𝐴 + 𝑏 ) = 0 ∧ ( 𝑏 + 𝐴 ) = 0 ) )
2		simpl	⊢ ( ( ( 𝐴 + 𝑏 ) = 0 ∧ ( 𝑏 + 𝐴 ) = 0 ) → ( 𝐴 + 𝑏 ) = 0 )
3	2	reximi	⊢ ( ∃ 𝑏 ∈ ℂ ( ( 𝐴 + 𝑏 ) = 0 ∧ ( 𝑏 + 𝐴 ) = 0 ) → ∃ 𝑏 ∈ ℂ ( 𝐴 + 𝑏 ) = 0 )
4	1 3	syl	⊢ ( 𝐴 ∈ ℂ → ∃ 𝑏 ∈ ℂ ( 𝐴 + 𝑏 ) = 0 )