This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.
Description: Distribution of multiplication over subtraction. Theorem I.5 of Apostol p. 18. (Contributed by NM, 18-Nov-2004)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | subdi | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 · ( 𝐵 − 𝐶 ) ) = ( ( 𝐴 · 𝐵 ) − ( 𝐴 · 𝐶 ) ) ) |
| 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 | 1 2 4 | adddid | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 · ( 𝐶 + ( 𝐵 − 𝐶 ) ) ) = ( ( 𝐴 · 𝐶 ) + ( 𝐴 · ( 𝐵 − 𝐶 ) ) ) ) |
| 6 | pncan3 | ⊢ ( ( 𝐶 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐶 + ( 𝐵 − 𝐶 ) ) = 𝐵 ) | |
| 7 | 6 | ancoms | ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐶 + ( 𝐵 − 𝐶 ) ) = 𝐵 ) |
| 8 | 7 | 3adant1 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐶 + ( 𝐵 − 𝐶 ) ) = 𝐵 ) |
| 9 | 8 | oveq2d | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 · ( 𝐶 + ( 𝐵 − 𝐶 ) ) ) = ( 𝐴 · 𝐵 ) ) |
| 10 | 5 9 | eqtr3d | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 · 𝐶 ) + ( 𝐴 · ( 𝐵 − 𝐶 ) ) ) = ( 𝐴 · 𝐵 ) ) |
| 11 | mulcl | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 · 𝐵 ) ∈ ℂ ) | |
| 12 | 11 | 3adant3 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 · 𝐵 ) ∈ ℂ ) |
| 13 | mulcl | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 · 𝐶 ) ∈ ℂ ) | |
| 14 | 13 | 3adant2 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 · 𝐶 ) ∈ ℂ ) |
| 15 | mulcl | ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 − 𝐶 ) ∈ ℂ ) → ( 𝐴 · ( 𝐵 − 𝐶 ) ) ∈ ℂ ) | |
| 16 | 3 15 | sylan2 | ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ) → ( 𝐴 · ( 𝐵 − 𝐶 ) ) ∈ ℂ ) |
| 17 | 16 | 3impb | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 · ( 𝐵 − 𝐶 ) ) ∈ ℂ ) |
| 18 | 12 14 17 | subaddd | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 · 𝐵 ) − ( 𝐴 · 𝐶 ) ) = ( 𝐴 · ( 𝐵 − 𝐶 ) ) ↔ ( ( 𝐴 · 𝐶 ) + ( 𝐴 · ( 𝐵 − 𝐶 ) ) ) = ( 𝐴 · 𝐵 ) ) ) |
| 19 | 10 18 | mpbird | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 · 𝐵 ) − ( 𝐴 · 𝐶 ) ) = ( 𝐴 · ( 𝐵 − 𝐶 ) ) ) |
| 20 | 19 | eqcomd | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 · ( 𝐵 − 𝐶 ) ) = ( ( 𝐴 · 𝐵 ) − ( 𝐴 · 𝐶 ) ) ) |