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Description: Swap the first and third terms in a double subtraction. (Contributed by Glauco Siliprandi, 11-Dec-2019)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | sub31 | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A - ( B - C ) ) = ( C - ( B - A ) ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | simp1 | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> A e. CC ) |
|
| 2 | simpr | |- ( ( B e. CC /\ C e. CC ) -> C e. CC ) |
|
| 3 | simpl | |- ( ( B e. CC /\ C e. CC ) -> B e. CC ) |
|
| 4 | 2 3 | subcld | |- ( ( B e. CC /\ C e. CC ) -> ( C - B ) e. CC ) |
| 5 | 4 | 3adant1 | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( C - B ) e. CC ) |
| 6 | 1 5 | addcomd | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A + ( C - B ) ) = ( ( C - B ) + A ) ) |
| 7 | subsub2 | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A - ( B - C ) ) = ( A + ( C - B ) ) ) |
|
| 8 | simp3 | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> C e. CC ) |
|
| 9 | simp2 | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> B e. CC ) |
|
| 10 | 8 9 1 | subsubd | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( C - ( B - A ) ) = ( ( C - B ) + A ) ) |
| 11 | 6 7 10 | 3eqtr4d | |- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A - ( B - C ) ) = ( C - ( B - A ) ) ) |