Theorem pldofph 41112

Description: Given, a,b c, d, "definition" for e, e is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)

Hypotheses

Ref	Expression
pldofph.1	|- ( ta <-> ( ( ch -> th ) /\ ( ph <-> ch ) /\ ( ( ph -> ps ) -> ( ps <-> th ) ) ) )
pldofph.2	|- ph
pldofph.3	|- ps
pldofph.4	|- ch
pldofph.5	|- th

Assertion

Ref	Expression
pldofph	|- ta

Proof of Theorem pldofph

Step	Hyp	Ref	Expression
1		pldofph.5	. . . 4 |- th
2	1	a1i 11	. . 3 |- ( ch -> th )
3		pldofph.2	. . . 4 |- ph
4		pldofph.4	. . . 4 |- ch
5	3, 4	2th 254	. . 3 |- ( ph <-> ch )
6		pldofph.3	. . . . 5 |- ps
7	6, 1	2th 254	. . . 4 |- ( ps <-> th )
8	7	a1i 11	. . 3 |- ( ( ph -> ps ) -> ( ps <-> th ) )
9	2, 5, 8	3pm3.2i 1239	. 2 |- ( ( ch -> th ) /\ ( ph <-> ch ) /\ ( ( ph -> ps ) -> ( ps <-> th ) ) )
10		pldofph.1	. . . 4 |- ( ta <-> ( ( ch -> th ) /\ ( ph <-> ch ) /\ ( ( ph -> ps ) -> ( ps <-> th ) ) ) )
11	10	bicomi 214	. . 3 |- ( ( ( ch -> th ) /\ ( ph <-> ch ) /\ ( ( ph -> ps ) -> ( ps <-> th ) ) ) <-> ta )
12	11	biimpi 206	. 2 |- ( ( ( ch -> th ) /\ ( ph <-> ch ) /\ ( ( ph -> ps ) -> ( ps <-> th ) ) ) -> ta )
13	9, 12	ax-mp 5	1 |- ta

Syntax hints:   -> wi 4   <-> wb 196   /\ w3a 1037

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 197  df-an 386  df-3an 1039

This theorem is referenced by:  plvcofph  41113  plvcofphax  41114