Description: Transfer two terms of a subtraction in an equality. (Contributed by Thierry Arnoux, 2-Feb-2020)
| Ref | Expression | ||
|---|---|---|---|
| Hypotheses | subeqxfrd.a | |- ( ph -> A e. CC ) |
|
| subeqxfrd.b | |- ( ph -> B e. CC ) |
||
| subeqxfrd.c | |- ( ph -> C e. CC ) |
||
| subeqxfrd.d | |- ( ph -> D e. CC ) |
||
| subeqxfrd.1 | |- ( ph -> ( A - B ) = ( C - D ) ) |
||
| Assertion | subeqxfrd | |- ( ph -> ( A - C ) = ( B - D ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | subeqxfrd.a | |- ( ph -> A e. CC ) |
|
| 2 | subeqxfrd.b | |- ( ph -> B e. CC ) |
|
| 3 | subeqxfrd.c | |- ( ph -> C e. CC ) |
|
| 4 | subeqxfrd.d | |- ( ph -> D e. CC ) |
|
| 5 | subeqxfrd.1 | |- ( ph -> ( A - B ) = ( C - D ) ) |
|
| 6 | 5 | oveq1d | |- ( ph -> ( ( A - B ) + ( B - C ) ) = ( ( C - D ) + ( B - C ) ) ) |
| 7 | 1 2 3 | npncand | |- ( ph -> ( ( A - B ) + ( B - C ) ) = ( A - C ) ) |
| 8 | 3 4 2 | npncan3d | |- ( ph -> ( ( C - D ) + ( B - C ) ) = ( B - D ) ) |
| 9 | 6 7 8 | 3eqtr3d | |- ( ph -> ( A - C ) = ( B - D ) ) |