Theorem sbequ2 864

Description: An equality theorem for substitution.

Assertion

Ref	Expression
sbequ2	⊢ (x = y → ([y / x]φ → φ))

Proof of Theorem sbequ2

Step	Hyp	Ref	Expression
1		pm3.26 256	. . 3 ⊢ (((x = y → φ) ∧ ∃x(x = y ∧ φ)) → (x = y → φ))
2	1	com12 13	. 2 ⊢ (x = y → (((x = y → φ) ∧ ∃x(x = y ∧ φ)) → φ))
3		df-sb 853	. 2 ⊢ ([y / x]φ ↔ ((x = y → φ) ∧ ∃x(x = y ∧ φ)))
4	2, 3	syl5ib 181	1 ⊢ (x = y → ([y / x]φ → φ))

Syntax hints:   → wi 2   ∧ wa 196  ∃wex 678   = weq 797  [wsb 852

This theorem is referenced by:  sbequ12 865  sb6y 872  hbsb3 875  sbequi 876  sbn1 880  sbi1 884  hbsb4 905  mo 1020  mopick 1054

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

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

metamath.org