Theorem sbco 1250

Description: A composition law for substitution.

Assertion

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

Proof of Theorem sbco

Step	Hyp	Ref	Expression
1		equsb2 1192	. . 3 |- [y / x]y = x
2		sbequ12 1179	. . . . 5 |- (y = x -> (ph <-> [x / y]ph))
3	2	bicomd 520	. . . 4 |- (y = x -> ([x / y]ph <-> ph))
4	3	sbimi 1171	. . 3 |- ([y / x]y = x -> [y / x]([x / y]ph <-> ph))
5	1, 4	ax-mp 7	. 2 |- [y / x]([x / y]ph <-> ph)
6		sbbi 1237	. 2 |- ([y / x]([x / y]ph <-> ph) <-> ([y / x][x / y]ph <-> [y / x]ph))
7	5, 6	mpbi 189	1 |- ([y / x][x / y]ph <-> [y / x]ph)

Syntax hints:   <-> wb 146  [wsbc 1168

This theorem is referenced by:  sbid2 1251  sbco3 1255  sb6rf 1258  sb9i 1261

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-10 964  ax-12 966  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-11o 1216

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 979  df-sb 1170

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