Theorem sb2 2513

Description: One direction of a simplified definition of substitution. The converse requires either a disjoint variable condition (sb6 2122) or a nonfreeness hypothesis (sb6f 2531). Usage of this theorem is discouraged because it depends on ax-13 2406. (Contributed by NM, 13-May-1993.) Revise df-sb 2100. (Revised by Wolf Lammen, 26-Jul-2023.) (New usage is discouraged.)

Assertion

Ref	Expression
sb2	⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜑)

Proof of Theorem sb2

Step	Hyp	Ref	Expression
1		pm2.27 43	. . . 4 ⊢ (𝑥 = 𝑦 → ((𝑥 = 𝑦 → 𝜑) → 𝜑))
2	1	al2imi 1848	. . 3 ⊢ (∀𝑥 𝑥 = 𝑦 → (∀𝑥(𝑥 = 𝑦 → 𝜑) → ∀𝑥𝜑))
3		stdpc4 2105	. . 3 ⊢ (∀𝑥𝜑 → [𝑦 / 𝑥]𝜑)
4	2, 3	syl6 36	. 2 ⊢ (∀𝑥 𝑥 = 𝑦 → (∀𝑥(𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜑))
5		sb4b 2509	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑)))
6	5	biimprd 251	. 2 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑥(𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜑))
7	4, 6	pm2.61i 184	1 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜑)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4  ∀wal 1568  [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:  hbsb2  2516  hbsb2a  2518  hbsb2e  2520  equsb1  2525  equsb2  2526  dfsb2  2527  sb6f  2531  sbeqal1  45141