Theorem sb1 2512

Description: One direction of a simplified definition of substitution. The converse requires either a disjoint variable condition (sb5 2313) or a nonfreeness hypothesis (sb5f 2532). Usage of this theorem is discouraged because it depends on ax-13 2406. Use the weaker sb1v 2124 when possible. (Contributed by NM, 13-May-1993.) Revise df-sb 2100. (Revised by Wolf Lammen, 21-Feb-2024.) (New usage is discouraged.)

Assertion

Ref	Expression
sb1	⊢ ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))

Proof of Theorem sb1

Step	Hyp	Ref	Expression
1		spsbe 2119	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → ∃𝑥𝜑)
2		pm3.2 475	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 → (𝑥 = 𝑦 ∧ 𝜑)))
3	2	aleximi 1865	. . 3 ⊢ (∀𝑥 𝑥 = 𝑦 → (∃𝑥𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
4	1, 3	syl5 35	. 2 ⊢ (∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
5		sb3b 2510	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
6	5	biimpd 232	. 2 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
7	4, 6	pm2.61i 184	1 ⊢ ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401  ∀wal 1568  ∃wex 1812  [wsb 2099

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2179  ax-12 2216  ax-13 2406

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817  df-sb 2100

This theorem is used by:  dfsb1  2515  sb4e  2519