Theorem sbequ1 1756

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		pm3.4 331	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜑))
2		19.8a 1578	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
3		df-sb 1751	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
4	1, 2, 3	sylanbrc 414	. 2 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑)
5	4	ex 114	1 ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))

Syntax hints:   → wi 4   ∧ wa 103  ∃wex 1480  [wsb 1750

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-4 1498

This theorem depends on definitions:  df-bi 116  df-sb 1751

This theorem is referenced by:  sbequ12  1759  sbequi  1827  sb6rf  1841  mo2n  2042  bj-bdfindes  13831  bj-findes  13863