Theorem sb4a 2514

Description: A version of one implication of sb4b 2509 that does not require a distinctor antecedent. Usage of this theorem is discouraged because it depends on ax-13 2406. Use the weaker sb4av 2282 when possible. (Contributed by NM, 2-Feb-2007.) Revise df-sb 2100. (Revised by Wolf Lammen, 28-Jul-2023.) (New usage is discouraged.)

Assertion

Ref	Expression
sb4a	⊢ ([𝑡 / 𝑥]∀𝑡𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜑))

Proof of Theorem sb4a

Step	Hyp	Ref	Expression
1		sbequ2 2287	. . . 4 ⊢ (𝑥 = 𝑡 → ([𝑡 / 𝑥]∀𝑡𝜑 → ∀𝑡𝜑))
2	1	sps 2224	. . 3 ⊢ (∀𝑥 𝑥 = 𝑡 → ([𝑡 / 𝑥]∀𝑡𝜑 → ∀𝑡𝜑))
3		axc11r 2402	. . . 4 ⊢ (∀𝑥 𝑥 = 𝑡 → (∀𝑡𝜑 → ∀𝑥𝜑))
4		ala1 1846	. . . 4 ⊢ (∀𝑥𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜑))
5	3, 4	syl6 36	. . 3 ⊢ (∀𝑥 𝑥 = 𝑡 → (∀𝑡𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜑)))
6	2, 5	syld 48	. 2 ⊢ (∀𝑥 𝑥 = 𝑡 → ([𝑡 / 𝑥]∀𝑡𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜑)))
7		sb4b 2509	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑡 → ([𝑡 / 𝑥]∀𝑡𝜑 ↔ ∀𝑥(𝑥 = 𝑡 → ∀𝑡𝜑)))
8		sp 2222	. . . . 5 ⊢ (∀𝑡𝜑 → 𝜑)
9	8	imim2i 17	. . . 4 ⊢ ((𝑥 = 𝑡 → ∀𝑡𝜑) → (𝑥 = 𝑡 → 𝜑))
10	9	alimi 1844	. . 3 ⊢ (∀𝑥(𝑥 = 𝑡 → ∀𝑡𝜑) → ∀𝑥(𝑥 = 𝑡 → 𝜑))
11	7, 10	biimtrdi 256	. 2 ⊢ (¬ ∀𝑥 𝑥 = 𝑡 → ([𝑡 / 𝑥]∀𝑡𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜑)))
12	6, 11	pm2.61i 184	1 ⊢ ([𝑡 / 𝑥]∀𝑡𝜑 → ∀𝑥(𝑥 = 𝑡 → 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:  hbsb2a  2518  sb6f  2531