Theorem eqsb3lemOLD 1988

Description: Lemma for eqsb3 1989.

Assertion

Ref	Expression
eqsb3lemOLD	|- ([x / y]y = A <-> x = A)

Distinct variable groups:   x,y   y,A

Proof of Theorem eqsb3lemOLD

Step	Hyp	Ref	Expression
1		equsb2 1562	. . . 4 |- [x / y]x = y
2		eqeq1 1890	. . . . 5 |- (x = y -> (x = A <-> y = A))
3	2	sbimi 1537	. . . 4 |- ([x / y]x = y -> [x / y](x = A <-> y = A))
4	1, 3	ax-mp 7	. . 3 |- [x / y](x = A <-> y = A)
5		sbbi 1609	. . 3 |- ([x / y](x = A <-> y = A) <-> ([x / y]x = A <-> [x / y]y = A))
6	4, 5	mpbi 206	. 2 |- ([x / y]x = A <-> [x / y]y = A)
7		ax-17 1317	. . 3 |- (x = A -> A.y x = A)
8	7	sbf 1551	. 2 |- ([x / y]x = A <-> x = A)
9	6, 8	bitr3i 192	1 |- ([x / y]y = A <-> x = A)

Syntax hints:   <-> wb 163   = wceq 1298  [wsbc 1534

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-10 1308  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-11o 1588  ax-ext 1865

This theorem depends on definitions:  df-bi 164  df-an 242  df-ex 1327  df-sb 1536  df-cleq 1877

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