Theorem sb2 2500

Description: One direction of a simplified definition of substitution. The converse requires either a disjoint variable condition (sb6 2089) or a non-freeness hypothesis (sb6f 2533). Usage of this theorem is discouraged because it depends on ax-13 2386. (Contributed by NM, 13-May-1993.) Revise df-sb 2066. (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 42	. . . 4 ⊢ (𝑥 = 𝑦 → ((𝑥 = 𝑦 → 𝜑) → 𝜑))
2	1	al2imi 1812	. . 3 ⊢ (∀𝑥 𝑥 = 𝑦 → (∀𝑥(𝑥 = 𝑦 → 𝜑) → ∀𝑥𝜑))
3		stdpc4 2069	. . 3 ⊢ (∀𝑥𝜑 → [𝑦 / 𝑥]𝜑)
4	2, 3	syl6 35	. 2 ⊢ (∀𝑥 𝑥 = 𝑦 → (∀𝑥(𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜑))
5		sb4b 2495	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑)))
6	5	biimprd 250	. 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)

Syntax hints:  ¬ wn 3   → wi 4  ∀wal 1531  [wsb 2065

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-10 2141  ax-12 2172  ax-13 2386

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1777  df-nf 1781  df-sb 2066

This theorem is referenced by:  sb3OLD  2501  hbsb2  2517  hbsb2a  2519  hbsb2e  2521  equsb1  2526  equsb2  2527  dfsb2  2528  sbequiOLD  2530  sb6f  2533  sbi1OLD  2538  sbeqal1  40723