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Theorem sbeqal1i 27010
Description: Suppose you know  x  =  y implies  x  =  z, assuming  x and  z are distinct. Then,  y  =  z. (Contributed by Andrew Salmon, 3-Jun-2011.)
Hypothesis
Ref Expression
sbeqal1i.1  |-  ( x  =  y  ->  x  =  z )
Assertion
Ref Expression
sbeqal1i  |-  y  =  z
Distinct variable group:    x, z

Proof of Theorem sbeqal1i
StepHypRef Expression
1 sbeqal1 27009 . 2  |-  ( A. x ( x  =  y  ->  x  =  z )  ->  y  =  z )
2 sbeqal1i.1 . 2  |-  ( x  =  y  ->  x  =  z )
31, 2mpg 1535 1  |-  y  =  z
Colors of variables: wff set class
Syntax hints:    -> wi 4    = wceq 1623
This theorem is referenced by:  sbeqal2i  27011
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1533  ax-5 1544  ax-17 1603  ax-9 1635  ax-8 1643  ax-6 1703  ax-7 1708  ax-11 1715  ax-12 1866
This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1310  df-ex 1529  df-nf 1532  df-sb 1630
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