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Metamath Proof Explorer

Theorem subsub23

Description: Swap subtrahend and result of subtraction. (Contributed by NM, 14-Dec-2007)

Ref Expression
Assertion subsub23 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) = 𝐶 ↔ ( 𝐴 − 𝐶 ) = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 addcom ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐶 ) = ( 𝐶 + 𝐵 ) )
2 1 3adant1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐶 ) = ( 𝐶 + 𝐵 ) )
3 2 eqeq1d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 + 𝐶 ) = 𝐴 ↔ ( 𝐶 + 𝐵 ) = 𝐴 ) )
4 subadd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) = 𝐶 ↔ ( 𝐵 + 𝐶 ) = 𝐴 ) )
5 subadd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) = 𝐵 ↔ ( 𝐶 + 𝐵 ) = 𝐴 ) )
6 5 3com23 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) = 𝐵 ↔ ( 𝐶 + 𝐵 ) = 𝐴 ) )
7 3 4 6 3bitr4d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) = 𝐶 ↔ ( 𝐴 − 𝐶 ) = 𝐵 ) )