Theorem sbidmOLD 1628

Description: An idempotent law for substitution.

Assertion

Ref	Expression
sbidmOLD	|- ([y / x][y / x]ph <-> [y / x]ph)

Proof of Theorem sbidmOLD

Step	Hyp	Ref	Expression
1		sbequ12 1545	. . . 4 |- (x = y -> ([y / x]ph <-> [y / x][y / x]ph))
2	1	bicomd 580	. . 3 |- (x = y -> ([y / x][y / x]ph <-> [y / x]ph))
3	2	a4s 1330	. 2 |- (A.x x = y -> ([y / x][y / x]ph <-> [y / x]ph))
4		hbnae 1507	. . 3 |- (-. A.x x = y -> A.x -. A.x x = y)
5		hbsb2 1597	. . 3 |- (-. A.x x = y -> ([y / x]ph -> A.x[y / x]ph))
6		biidd 188	. . . 4 |- (x = y -> ([y / x]ph <-> [y / x]ph))
7	6	a1i 8	. . 3 |- (-. A.x x = y -> (x = y -> ([y / x]ph <-> [y / x]ph)))
8	4, 5, 7	sbied 1563	. 2 |- (-. A.x x = y -> ([y / x][y / x]ph <-> [y / x]ph))
9	3, 8	pm2.61i 140	1 |- ([y / x][y / x]ph <-> [y / x]ph)

Syntax hints:  -. wn 2   -> wi 3   <-> wb 163  A.wal 1296  [wsbc 1534

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-10 1308  ax-12 1310  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-11o 1588

This theorem depends on definitions:  df-bi 164  df-an 242  df-ex 1327  df-sb 1536

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