Theorem sbequ2 1692

Description: An equality theorem for substitution. (Contributed by NM, 5-Aug-1993.)

Assertion

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

Proof of Theorem sbequ2

Step	Hyp	Ref	Expression
1		df-sb 1686	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
2		simpl 107	. . 3 ⊢ (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → (𝑥 = 𝑦 → 𝜑))
3	2	com12 30	. 2 ⊢ (𝑥 = 𝑦 → (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → 𝜑))
4	1, 3	syl5bi 150	1 ⊢ (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 → 𝜑))

Syntax hints:   → wi 4   ∧ wa 102  ∃wex 1421  [wsb 1685

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104

This theorem depends on definitions:  df-bi 115  df-sb 1686

This theorem is referenced by:  stdpc7  1693  sbequ12  1694  sbequi  1760  mo23  1982  mopick  2019