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Metamath Proof Explorer

Theorem subcan2

Description: Cancellation law for subtraction. (Contributed by NM, 8-Feb-2005)

Ref Expression
Assertion subcan2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) = ( 𝐵 − 𝐶 ) ↔ 𝐴 = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 simp1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐴 ∈ ℂ )
2 simp3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐶 ∈ ℂ )
3 subcl ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 − 𝐶 ) ∈ ℂ )
4 3 3adant1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 − 𝐶 ) ∈ ℂ )
5 subadd2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ ( 𝐵 − 𝐶 ) ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) = ( 𝐵 − 𝐶 ) ↔ ( ( 𝐵 − 𝐶 ) + 𝐶 ) = 𝐴 ) )
6 1 2 4 5 syl3anc ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) = ( 𝐵 − 𝐶 ) ↔ ( ( 𝐵 − 𝐶 ) + 𝐶 ) = 𝐴 ) )
7 npcan ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 − 𝐶 ) + 𝐶 ) = 𝐵 )
8 7 3adant1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 − 𝐶 ) + 𝐶 ) = 𝐵 )
9 8 eqeq1d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐵 − 𝐶 ) + 𝐶 ) = 𝐴 ↔ 𝐵 = 𝐴 ) )
10 eqcom ⊢ ( 𝐵 = 𝐴 ↔ 𝐴 = 𝐵 )
11 9 10 bitrdi ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐵 − 𝐶 ) + 𝐶 ) = 𝐴 ↔ 𝐴 = 𝐵 ) )
12 6 11 bitrd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) = ( 𝐵 − 𝐶 ) ↔ 𝐴 = 𝐵 ) )