Description: Eliminate an antecedent implied by each side of a biconditional. Variant of pm5.21ndd . (Contributed by NM, 20-Nov-2005) (Proof shortened by Wolf Lammen, 4-Nov-2013)
Ref | Expression | ||
---|---|---|---|
Hypotheses | pm5.21nd.1 | |- ( ( ph /\ ps ) -> th ) |
|
pm5.21nd.2 | |- ( ( ph /\ ch ) -> th ) |
||
pm5.21nd.3 | |- ( th -> ( ps <-> ch ) ) |
||
Assertion | pm5.21nd | |- ( ph -> ( ps <-> ch ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | pm5.21nd.1 | |- ( ( ph /\ ps ) -> th ) |
|
2 | pm5.21nd.2 | |- ( ( ph /\ ch ) -> th ) |
|
3 | pm5.21nd.3 | |- ( th -> ( ps <-> ch ) ) |
|
4 | 1 | ex | |- ( ph -> ( ps -> th ) ) |
5 | 2 | ex | |- ( ph -> ( ch -> th ) ) |
6 | 3 | a1i | |- ( ph -> ( th -> ( ps <-> ch ) ) ) |
7 | 4 5 6 | pm5.21ndd | |- ( ph -> ( ps <-> ch ) ) |