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Description: Distribution of multiplication over subtraction. Theorem I.5 of Apostol p. 18. (Contributed by NM, 18-Nov-2004)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | subdi | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A x. ( B - C ) ) = ( ( A x. B ) - ( A x. C ) ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | simp1 | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> A e. CC ) |
|
| 2 | simp3 | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> C e. CC ) |
|
| 3 | subcl | |- ( ( B e. CC /\ C e. CC ) -> ( B - C ) e. CC ) |
|
| 4 | 3 | 3adant1 | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( B - C ) e. CC ) |
| 5 | 1 2 4 | adddid | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A x. ( C + ( B - C ) ) ) = ( ( A x. C ) + ( A x. ( B - C ) ) ) ) |
| 6 | pncan3 | |- ( ( C e. CC /\ B e. CC ) -> ( C + ( B - C ) ) = B ) |
|
| 7 | 6 | ancoms | |- ( ( B e. CC /\ C e. CC ) -> ( C + ( B - C ) ) = B ) |
| 8 | 7 | 3adant1 | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( C + ( B - C ) ) = B ) |
| 9 | 8 | oveq2d | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A x. ( C + ( B - C ) ) ) = ( A x. B ) ) |
| 10 | 5 9 | eqtr3d | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A x. C ) + ( A x. ( B - C ) ) ) = ( A x. B ) ) |
| 11 | mulcl | |- ( ( A e. CC /\ B e. CC ) -> ( A x. B ) e. CC ) |
|
| 12 | 11 | 3adant3 | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A x. B ) e. CC ) |
| 13 | mulcl | |- ( ( A e. CC /\ C e. CC ) -> ( A x. C ) e. CC ) |
|
| 14 | 13 | 3adant2 | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A x. C ) e. CC ) |
| 15 | mulcl | |- ( ( A e. CC /\ ( B - C ) e. CC ) -> ( A x. ( B - C ) ) e. CC ) |
|
| 16 | 3 15 | sylan2 | |- ( ( A e. CC /\ ( B e. CC /\ C e. CC ) ) -> ( A x. ( B - C ) ) e. CC ) |
| 17 | 16 | 3impb | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A x. ( B - C ) ) e. CC ) |
| 18 | 12 14 17 | subaddd | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( ( A x. B ) - ( A x. C ) ) = ( A x. ( B - C ) ) <-> ( ( A x. C ) + ( A x. ( B - C ) ) ) = ( A x. B ) ) ) |
| 19 | 10 18 | mpbird | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A x. B ) - ( A x. C ) ) = ( A x. ( B - C ) ) ) |
| 20 | 19 | eqcomd | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A x. ( B - C ) ) = ( ( A x. B ) - ( A x. C ) ) ) |