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Description: Cancellation law for addition. (Contributed by NM, 30-Jul-2004) (Revised by Scott Fenton, 3-Jan-2013)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | addcan2 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) = ( 𝐵 + 𝐶 ) ↔ 𝐴 = 𝐵 ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | cnegex | ⊢ ( 𝐶 ∈ ℂ → ∃ 𝑥 ∈ ℂ ( 𝐶 + 𝑥 ) = 0 ) | |
| 2 | 1 | 3ad2ant3 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ∃ 𝑥 ∈ ℂ ( 𝐶 + 𝑥 ) = 0 ) |
| 3 | oveq1 | ⊢ ( ( 𝐴 + 𝐶 ) = ( 𝐵 + 𝐶 ) → ( ( 𝐴 + 𝐶 ) + 𝑥 ) = ( ( 𝐵 + 𝐶 ) + 𝑥 ) ) | |
| 4 | simpl1 | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → 𝐴 ∈ ℂ ) | |
| 5 | simpl3 | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → 𝐶 ∈ ℂ ) | |
| 6 | simprl | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → 𝑥 ∈ ℂ ) | |
| 7 | 4 5 6 | addassd | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → ( ( 𝐴 + 𝐶 ) + 𝑥 ) = ( 𝐴 + ( 𝐶 + 𝑥 ) ) ) |
| 8 | simprr | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → ( 𝐶 + 𝑥 ) = 0 ) | |
| 9 | 8 | oveq2d | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → ( 𝐴 + ( 𝐶 + 𝑥 ) ) = ( 𝐴 + 0 ) ) |
| 10 | addrid | ⊢ ( 𝐴 ∈ ℂ → ( 𝐴 + 0 ) = 𝐴 ) | |
| 11 | 4 10 | syl | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → ( 𝐴 + 0 ) = 𝐴 ) |
| 12 | 7 9 11 | 3eqtrd | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → ( ( 𝐴 + 𝐶 ) + 𝑥 ) = 𝐴 ) |
| 13 | simpl2 | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → 𝐵 ∈ ℂ ) | |
| 14 | 13 5 6 | addassd | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → ( ( 𝐵 + 𝐶 ) + 𝑥 ) = ( 𝐵 + ( 𝐶 + 𝑥 ) ) ) |
| 15 | 8 | oveq2d | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → ( 𝐵 + ( 𝐶 + 𝑥 ) ) = ( 𝐵 + 0 ) ) |
| 16 | addrid | ⊢ ( 𝐵 ∈ ℂ → ( 𝐵 + 0 ) = 𝐵 ) | |
| 17 | 13 16 | syl | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → ( 𝐵 + 0 ) = 𝐵 ) |
| 18 | 14 15 17 | 3eqtrd | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → ( ( 𝐵 + 𝐶 ) + 𝑥 ) = 𝐵 ) |
| 19 | 12 18 | eqeq12d | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → ( ( ( 𝐴 + 𝐶 ) + 𝑥 ) = ( ( 𝐵 + 𝐶 ) + 𝑥 ) ↔ 𝐴 = 𝐵 ) ) |
| 20 | 3 19 | imbitrid | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → ( ( 𝐴 + 𝐶 ) = ( 𝐵 + 𝐶 ) → 𝐴 = 𝐵 ) ) |
| 21 | oveq1 | ⊢ ( 𝐴 = 𝐵 → ( 𝐴 + 𝐶 ) = ( 𝐵 + 𝐶 ) ) | |
| 22 | 20 21 | impbid1 | ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝑥 ∈ ℂ ∧ ( 𝐶 + 𝑥 ) = 0 ) ) → ( ( 𝐴 + 𝐶 ) = ( 𝐵 + 𝐶 ) ↔ 𝐴 = 𝐵 ) ) |
| 23 | 2 22 | rexlimddv | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) = ( 𝐵 + 𝐶 ) ↔ 𝐴 = 𝐵 ) ) |