HomeHome Metamath Proof Explorer < Previous   Next >
Related theorems
Unicode version

Theorem sbal 1329
Description: Move universal quantifier in and out of substitution.
Assertion
Ref Expression
sbal |- ([z / y]A.xph <-> A.x[z / y]ph)
Distinct variable groups:   x,y   x,z

Proof of Theorem sbal
StepHypRef Expression
1 a16gb 1259 . . . . 5 |- (A.x x = z -> (ph <-> A.xph))
21sbimi 1156 . . . 4 |- ([z / y]A.x x = z -> [z / y](ph <-> A.xph))
3 sbequ5 1173 . . . 4 |- ([z / y]A.x x = z <-> A.x x = z)
4 sbbi 1223 . . . 4 |- ([z / y](ph <-> A.xph) <-> ([z / y]ph <-> [z / y]A.xph))
52, 3, 43imtr3 218 . . 3 |- (A.x x = z -> ([z / y]ph <-> [z / y]A.xph))
6 a16gb 1259 . . 3 |- (A.x x = z -> ([z / y]ph <-> A.x[z / y]ph))
75, 6bitr3d 528 . 2 |- (A.x x = z -> ([z / y]A.xph <-> A.x[z / y]ph))
8 sbal1 1328 . 2 |- (-. A.x x = z -> ([z / y]A.xph <-> A.x[z / y]ph))
97, 8pm2.61i 126 1 |- ([z / y]A.xph <-> A.x[z / y]ph)
Colors of variables: wff set class
Syntax hints:   <-> wb 146  A.wal 950  [wsbc 1153
This theorem is referenced by:  sbex 1330  sbalv 1331  sbabel 1560  sbcalg 1945
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-16 1194  ax-11o 1202
This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 957  df-sb 1155
Copyright terms: Public domain