Metamath Proof Explorer
Description: Given, a,b c, d, "definition" for e, e is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020)
|
|
Ref |
Expression |
|
Hypotheses |
pldofph.1 |
|- ( ta <-> ( ( ch -> th ) /\ ( ph <-> ch ) /\ ( ( ph -> ps ) -> ( ps <-> th ) ) ) ) |
|
|
pldofph.2 |
|- ph |
|
|
pldofph.3 |
|- ps |
|
|
pldofph.4 |
|- ch |
|
|
pldofph.5 |
|- th |
|
Assertion |
pldofph |
|- ta |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
pldofph.1 |
|- ( ta <-> ( ( ch -> th ) /\ ( ph <-> ch ) /\ ( ( ph -> ps ) -> ( ps <-> th ) ) ) ) |
2 |
|
pldofph.2 |
|- ph |
3 |
|
pldofph.3 |
|- ps |
4 |
|
pldofph.4 |
|- ch |
5 |
|
pldofph.5 |
|- th |
6 |
5
|
a1i |
|- ( ch -> th ) |
7 |
2 4
|
2th |
|- ( ph <-> ch ) |
8 |
3 5
|
2th |
|- ( ps <-> th ) |
9 |
8
|
a1i |
|- ( ( ph -> ps ) -> ( ps <-> th ) ) |
10 |
6 7 9
|
3pm3.2i |
|- ( ( ch -> th ) /\ ( ph <-> ch ) /\ ( ( ph -> ps ) -> ( ps <-> th ) ) ) |
11 |
1
|
bicomi |
|- ( ( ( ch -> th ) /\ ( ph <-> ch ) /\ ( ( ph -> ps ) -> ( ps <-> th ) ) ) <-> ta ) |
12 |
11
|
biimpi |
|- ( ( ( ch -> th ) /\ ( ph <-> ch ) /\ ( ( ph -> ps ) -> ( ps <-> th ) ) ) -> ta ) |
13 |
10 12
|
ax-mp |
|- ta |