Theorem sbequ1 2110

Description: An equality theorem for substitution. (Contributed by NM, 16-May-1993.)

Assertion

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

Proof of Theorem sbequ1

Step	Hyp	Ref	Expression
1		pm3.4 584	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜑))
2		19.8a 2052	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
3		df-sb 1881	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
4	1, 2, 3	sylanbrc 698	. 2 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑)
5	4	ex 450	1 ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))

Syntax hints:   → wi 4   ∧ wa 384  ∃wex 1704  [wsb 1880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-12 2047

This theorem depends on definitions:  df-bi 197  df-an 386  df-ex 1705  df-sb 1881

This theorem is referenced by:  sbequ12  2111  dfsb2  2373  sbequi  2375  sbi1  2392  2eu6  2558  sb5ALT  38731  2pm13.193  38768  2pm13.193VD  39139  sb5ALTVD  39149