Theorem sbidm 1254

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 1181	. . . 4 |- (x = y -> ([y / x]ph <-> [y / x][y / x]ph))
2	1	bicomd 521	. . 3 |- (x = y -> ([y / x][y / x]ph <-> [y / x]ph))
3	2	a4s 984	. 2 |- (A.x x = y -> ([y / x][y / x]ph <-> [y / x]ph))
4		hbnae 1147	. . 3 |- (-. A.x x = y -> A.x -. A.x x = y)
5		hbsb2 1227	. . 3 |- (-. A.x x = y -> ([y / x]ph -> A.x[y / x]ph))
6		pm4.2d 171	. . . 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 1195	. 2 |- (-. A.x x = y -> ([y / x][y / x]ph <-> [y / x]ph))
9	3, 8	pm2.61i 126	1 |- ([y / x][y / x]ph <-> [y / x]ph)

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146  A.wal 954  [wsbc 1170

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-10 966  ax-12 968  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-11o 1218

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172

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