Theorem sbequ1 863

Description: An equality theorem for substitution.

Assertion

Ref	Expression
sbequ1	⊢ (x = y → (φ → [y / x]φ))

Proof of Theorem sbequ1

Step	Hyp	Ref	Expression
1		pm3.4 266	. . . 4 ⊢ ((x = y ∧ φ) → (x = y → φ))
2		19.8a 712	. . . 4 ⊢ ((x = y ∧ φ) → ∃x(x = y ∧ φ))
3	1, 2	jca 236	. . 3 ⊢ ((x = y ∧ φ) → ((x = y → φ) ∧ ∃x(x = y ∧ φ)))
4		df-sb 853	. . 3 ⊢ ([y / x]φ ↔ ((x = y → φ) ∧ ∃x(x = y ∧ φ)))
5	3, 4	sylibr 175	. 2 ⊢ ((x = y ∧ φ) → [y / x]φ)
6	5	exp 291	1 ⊢ (x = y → (φ → [y / x]φ))

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

This theorem is referenced by:  sbequ12 865  hbsb3 875  sbequi 876  sbn2 881  sbi1 884  hbsb4 905  sb5f1 917  mo 1020

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673

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

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