Theorem sbidm 1256

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 1183	. . . 4 |- (x = y -> ([y / x]ph <-> [y / x][y / x]ph))
2	1	bicomd 523	. . 3 |- (x = y -> ([y / x][y / x]ph <-> [y / x]ph))
3	2	a4s 986	. 2 |- (A.x x = y -> ([y / x][y / x]ph <-> [y / x]ph))
4		hbnae 1149	. . 3 |- (-. A.x x = y -> A.x -. A.x x = y)
5		hbsb2 1229	. . 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 1197	. 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 956  [wsbc 1172

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-10 968  ax-12 970  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-11o 1220

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-sb 1174

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