Theorem sbequ1 1691

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

Assertion

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

Proof of Theorem sbequ1

Step	Hyp	Ref	Expression
1		pm3.4 326	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜑))
2		19.8a 1522	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
3		df-sb 1686	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
4	1, 2, 3	sylanbrc 408	. 2 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑)
5	4	ex 113	1 ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]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  ax-ia2 105  ax-ia3 106  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-4 1440

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

This theorem is referenced by:  sbequ12  1694  sbequi  1760  sb6rf  1774  mo2n  1969  bj-bdfindes  10744  bj-findes  10776