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Description: Partial products algorithm for two digit multiplication. (Contributed by Steven Nguyen, 10-Dec-2022)
| Ref | Expression | ||
|---|---|---|---|
| Hypotheses | decpmulnc.a | |- A e. NN0 |
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| decpmulnc.b | |- B e. NN0 |
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| decpmulnc.c | |- C e. NN0 |
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| decpmulnc.d | |- D e. NN0 |
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| decpmulnc.1 | |- ( A x. C ) = E |
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| decpmulnc.2 | |- ( ( A x. D ) + ( B x. C ) ) = F |
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| decpmul.3 | |- ( B x. D ) = ; G H |
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| decpmul.4 | |- ( ; E G + F ) = I |
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| decpmul.g | |- G e. NN0 |
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| decpmul.h | |- H e. NN0 |
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| Assertion | decpmul | |- ( ; A B x. ; C D ) = ; I H |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | decpmulnc.a | |- A e. NN0 |
|
| 2 | decpmulnc.b | |- B e. NN0 |
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| 3 | decpmulnc.c | |- C e. NN0 |
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| 4 | decpmulnc.d | |- D e. NN0 |
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| 5 | decpmulnc.1 | |- ( A x. C ) = E |
|
| 6 | decpmulnc.2 | |- ( ( A x. D ) + ( B x. C ) ) = F |
|
| 7 | decpmul.3 | |- ( B x. D ) = ; G H |
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| 8 | decpmul.4 | |- ( ; E G + F ) = I |
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| 9 | decpmul.g | |- G e. NN0 |
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| 10 | decpmul.h | |- H e. NN0 |
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| 11 | 1 2 3 4 5 6 7 | decpmulnc | |- ( ; A B x. ; C D ) = ; ; E F ; G H |
| 12 | dfdec10 | |- ; ; E F ; G H = ( ( ; 1 0 x. ; E F ) + ; G H ) |
|
| 13 | 1 3 | nn0mulcli | |- ( A x. C ) e. NN0 |
| 14 | 5 13 | eqeltrri | |- E e. NN0 |
| 15 | 2 3 | nn0mulcli | |- ( B x. C ) e. NN0 |
| 16 | 1 4 15 | numcl | |- ( ( A x. D ) + ( B x. C ) ) e. NN0 |
| 17 | 6 16 | eqeltrri | |- F e. NN0 |
| 18 | 14 17 | deccl | |- ; E F e. NN0 |
| 19 | 0nn0 | |- 0 e. NN0 |
|
| 20 | 18 | dec0u | |- ( ; 1 0 x. ; E F ) = ; ; E F 0 |
| 21 | eqid | |- ; G H = ; G H |
|
| 22 | 14 17 9 | decaddcom | |- ( ; E F + G ) = ( ; E G + F ) |
| 23 | 22 8 | eqtri | |- ( ; E F + G ) = I |
| 24 | 10 | nn0cni | |- H e. CC |
| 25 | 24 | addlidi | |- ( 0 + H ) = H |
| 26 | 18 19 9 10 20 21 23 25 | decadd | |- ( ( ; 1 0 x. ; E F ) + ; G H ) = ; I H |
| 27 | 11 12 26 | 3eqtri | |- ( ; A B x. ; C D ) = ; I H |