Theorem sbequ1 1768

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		pm3.4 333	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜑))
2		19.8a 1590	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
3		df-sb 1763	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
4	1, 2, 3	sylanbrc 417	. 2 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑)
5	4	ex 115	1 ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))

Syntax hints:   → wi 4   ∧ wa 104  ∃wex 1492  [wsb 1762

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-4 1510

This theorem depends on definitions:  df-bi 117  df-sb 1763

This theorem is referenced by:  sbequ12  1771  sbequi  1839  sb6rf  1853  mo2n  2054  bj-bdfindes  14583  bj-findes  14615