Theorem dral1 1150

Description: Formula-building lemma for use with the Distinctor Reduction Theorem. Part of Theorem 9.4 of [Megill] p. 448 (p. 16 of preprint).

Hypothesis

Ref	Expression
dral1.1	|- (A.x x = y -> (ph <-> ps))

Assertion

Ref	Expression
dral1	|- (A.x x = y -> (A.xph <-> A.yps))

Proof of Theorem dral1

Step	Hyp	Ref	Expression
1		hbae 1141	. . . 4 |- (A.x x = y -> A.xA.x x = y)
2		dral1.1	. . . . 5 |- (A.x x = y -> (ph <-> ps))
3	2	biimpd 153	. . . 4 |- (A.x x = y -> (ph -> ps))
4	1, 3	19.20d 993	. . 3 |- (A.x x = y -> (A.xph -> A.xps))
5		ax-10o 1136	. . 3 |- (A.x x = y -> (A.xps -> A.yps))
6	4, 5	syld 27	. 2 |- (A.x x = y -> (A.xph -> A.yps))
7		hbae 1141	. . . 4 |- (A.x x = y -> A.yA.x x = y)
8	2	biimprd 154	. . . 4 |- (A.x x = y -> (ps -> ph))
9	7, 8	19.20d 993	. . 3 |- (A.x x = y -> (A.yps -> A.yph))
10		ax-10o 1136	. . . 4 |- (A.y y = x -> (A.yph -> A.xph))
11	10	alequcoms 1139	. . 3 |- (A.x x = y -> (A.yph -> A.xph))
12	9, 11	syld 27	. 2 |- (A.x x = y -> (A.yps -> A.xph))
13	6, 12	impbid 514	1 |- (A.x x = y -> (A.xph <-> A.yps))

Syntax hints:   -> wi 3   <-> wb 146  A.wal 951   = wceq 953

This theorem is referenced by:  drex1 1152  ax11 1214  hbsb4 1243  sb9i 1258  a16g 1271  ax11indalem 1361  ax11inda2ALT 1362  ralcom2 1768  axpownd 4925

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 959  ax-gen 960  ax-10 963  ax-12 965  ax-4 970  ax-5o 972  ax-10o 1136

This theorem depends on definitions:  df-bi 147  df-an 225

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