Theorem sbequ8 902

Description: Elimination of equality from antecedent after substitution.

Assertion

Ref	Expression
sbequ8	⊢ ([y / x]φ ↔ [y / x](x = y → φ))

Proof of Theorem sbequ8

Step	Hyp	Ref	Expression
1		sbeq1 900	. . 3 ⊢ [y / x]x = y
2	1	a1bi 172	. 2 ⊢ ([y / x]φ ↔ ([y / x]x = y → [y / x]φ))
3		sbim 886	. 2 ⊢ ([y / x](x = y → φ) ↔ ([y / x]x = y → [y / x]φ))
4	2, 3	bitr4 154	1 ⊢ ([y / x]φ ↔ [y / x](x = y → φ))

Syntax hints:   → wi 2   ↔ wb 127   = weq 797  [wsb 852

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

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