Description: Eliminate an antecedent implied by each side of a biconditional, deduction version. (Contributed by Paul Chapman, 21-Nov-2012) (Proof shortened by Wolf Lammen, 6-Oct-2013)
| Ref | Expression | ||
|---|---|---|---|
| Hypotheses | pm5.21ndd.1 | |- ( ph -> ( ch -> ps ) ) |
|
| pm5.21ndd.2 | |- ( ph -> ( th -> ps ) ) |
||
| pm5.21ndd.3 | |- ( ph -> ( ps -> ( ch <-> th ) ) ) |
||
| Assertion | pm5.21ndd | |- ( ph -> ( ch <-> th ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | pm5.21ndd.1 | |- ( ph -> ( ch -> ps ) ) |
|
| 2 | pm5.21ndd.2 | |- ( ph -> ( th -> ps ) ) |
|
| 3 | pm5.21ndd.3 | |- ( ph -> ( ps -> ( ch <-> th ) ) ) |
|
| 4 | 1 | con3d | |- ( ph -> ( -. ps -> -. ch ) ) |
| 5 | 2 | con3d | |- ( ph -> ( -. ps -> -. th ) ) |
| 6 | pm5.21im | |- ( -. ch -> ( -. th -> ( ch <-> th ) ) ) |
|
| 7 | 4 5 6 | syl6c | |- ( ph -> ( -. ps -> ( ch <-> th ) ) ) |
| 8 | 3 7 | pm2.61d | |- ( ph -> ( ch <-> th ) ) |