Theorem sbeqal1i 16356

Description: Suppose you know x = y implies x = z, assuming x and z are distinct. Then, y = z.

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 16355	. 2 |- (A.x(x = y -> x = z) -> y = z)
2		sbeqal1i.1	. 2 |- (x = y -> x = z)
3	1, 2	mpg 1332	1 |- y = z

Syntax hints:   -> wi 3   = wceq 1298

This theorem is referenced by:  sbeqal2i 16357

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-8 1306  ax-9 1307  ax-10 1308  ax-11 1309  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-11o 1588

This theorem depends on definitions:  df-bi 164  df-or 241  df-an 242  df-ex 1327  df-sb 1536

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