Theorem equsb3lem 2101

Description: Lemma for equsb3 2102. (Contributed by Raph Levien and FL, 4-Dec-2005.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)

Assertion

Ref	Expression
equsb3lem	⊢ ([y / x]x = z ↔ y = z)

Distinct variable groups:   x,z   x,y

Proof of Theorem equsb3lem

Step	Hyp	Ref	Expression
1		nfv 1619	. 2 ⊢ Ⅎx y = z
2		equequ1 1684	. 2 ⊢ (x = y → (x = z ↔ y = z))
3	1, 2	sbie 2038	1 ⊢ ([y / x]x = z ↔ y = z)

Syntax hints:   ↔ wb 176  [wsb 1648

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is referenced by:  equsb3  2102