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Metamath Proof Explorer

Theorem subsub23d

Description: Swap subtrahend and result of subtraction. (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Hypotheses subsub23d.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
subsub23d.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
subsub23d.3 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
Assertion subsub23d ( 𝜑 → ( ( 𝐴 − 𝐵 ) = 𝐶 ↔ ( 𝐴 − 𝐶 ) = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 subsub23d.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 subsub23d.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3 subsub23d.3 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
4 subsub23 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) = 𝐶 ↔ ( 𝐴 − 𝐶 ) = 𝐵 ) )
5 1 2 3 4 syl3anc ⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) = 𝐶 ↔ ( 𝐴 − 𝐶 ) = 𝐵 ) )