Theorem sbidm 912

Description: An idempotent law for substitution.

Assertion

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

Proof of Theorem sbidm

Step	Hyp	Ref	Expression
1		sbequ12 865	. . . 4 |- (x = y -> ([y / x]ph <-> [y / x][y / x]ph))
2	1	bicomd 399	. . 3 |- (x = y -> ([y / x][y / x]ph <-> [y / x]ph))
3	2	a4s 682	. 2 |- (A.x x = y -> ([y / x][y / x]ph <-> [y / x]ph))
4		eq6 826	. . 3 |- (-. A.x x = y -> A.x -. A.x x = y)
5		hbsb2 873	. . 3 |- (-. A.x x = y -> ([y / x]ph -> A.x[y / x]ph))
6		pm4.2i 149	. . . 4 |- (x = y -> ([y / x]ph <-> [y / x]ph))
7	6	a1i 7	. . 3 |- (-. A.x x = y -> (x = y -> ([y / x]ph <-> [y / x]ph)))
8	4, 5, 7	sbied 903	. 2 |- (-. A.x x = y -> ([y / x][y / x]ph <-> [y / x]ph))
9	3, 8	pm2.61i 110	1 |- ([y / x][y / x]ph <-> [y / x]ph)

Syntax hints:  -. wn 1   -> wi 2   <-> wb 127  A.wal 672   = weq 797  [wsb 852

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

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