Theorem sbidm 1238

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 1164	. . . 4 |- (x = y -> ([y / x]ph <-> [y / x][y / x]ph))
2	1	bicomd 519	. . 3 |- (x = y -> ([y / x][y / x]ph <-> [y / x]ph))
3	2	a4s 960	. 2 |- (A.x x = y -> ([y / x][y / x]ph <-> [y / x]ph))
4		hbnae 1130	. . 3 |- (-. A.x x = y -> A.x -. A.x x = y)
5		hbsb2 1211	. . 3 |- (-. A.x x = y -> ([y / x]ph -> A.x[y / x]ph))
6		pm4.2i 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 1178	. 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 950  [wsbc 1153

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-4 951  ax-5 952  ax-6 953  ax-7 954  ax-gen 955  ax-8 1101  ax-9 1102  ax-10 1103  ax-12 1104  ax-11o 1202

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 957  df-sb 1155

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