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Metamath Proof Explorer

Theorem subval

Description: Value of subtraction, which is the (unique) element x such that B + x = A . (Contributed by NM, 4-Aug-2007) (Revised by Mario Carneiro, 2-Nov-2013)

		Ref	Expression
	Assertion	subval	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 − 𝐵 ) = ( ℩ 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		eqeq2	⊢ ( 𝑦 = 𝐴 → ( ( 𝑧 + 𝑥 ) = 𝑦 ↔ ( 𝑧 + 𝑥 ) = 𝐴 ) )
2	1	riotabidv	⊢ ( 𝑦 = 𝐴 → ( ℩ 𝑥 ∈ ℂ ( 𝑧 + 𝑥 ) = 𝑦 ) = ( ℩ 𝑥 ∈ ℂ ( 𝑧 + 𝑥 ) = 𝐴 ) )
3		oveq1	⊢ ( 𝑧 = 𝐵 → ( 𝑧 + 𝑥 ) = ( 𝐵 + 𝑥 ) )
4	3	eqeq1d	⊢ ( 𝑧 = 𝐵 → ( ( 𝑧 + 𝑥 ) = 𝐴 ↔ ( 𝐵 + 𝑥 ) = 𝐴 ) )
5	4	riotabidv	⊢ ( 𝑧 = 𝐵 → ( ℩ 𝑥 ∈ ℂ ( 𝑧 + 𝑥 ) = 𝐴 ) = ( ℩ 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) )
6		df-sub	⊢ − = ( 𝑦 ∈ ℂ , 𝑧 ∈ ℂ ↦ ( ℩ 𝑥 ∈ ℂ ( 𝑧 + 𝑥 ) = 𝑦 ) )
7		riotaex	⊢ ( ℩ 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) ∈ V
8	2 5 6 7	ovmpo	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 − 𝐵 ) = ( ℩ 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) )