Theorem sbeqal1i 38599

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

Step	Hyp	Ref	Expression
1		sbeqal1 38598	. 2 |- ( A. x ( x = y -> x = z ) -> y = z )
2		sbeqal1i.1	. 2 |- ( x = y -> x = z )
3	1, 2	mpg 1724	1 |- y = z

Syntax hints:   -> wi 4

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-12 2047  ax-13 2246

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1705  df-nf 1710  df-sb 1881

This theorem is referenced by:  sbeqal2i  38600