Theorem sb4a 2509

Description: A version of one implication of sb4b 2499 that does not require a distinctor antecedent. Usage of this theorem is discouraged because it depends on ax-13 2390. Use the weaker sb4av 2244 when possible. (Contributed by NM, 2-Feb-2007.) Revise df-sb 2070. (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 2250	. . . 4 ⊢ (𝑥 = 𝑡 → ([𝑡 / 𝑥]∀𝑡𝜑 → ∀𝑡𝜑))
2	1	sps 2184	. . 3 ⊢ (∀𝑥 𝑥 = 𝑡 → ([𝑡 / 𝑥]∀𝑡𝜑 → ∀𝑡𝜑))
3		axc11r 2386	. . . 4 ⊢ (∀𝑥 𝑥 = 𝑡 → (∀𝑡𝜑 → ∀𝑥𝜑))
4		ala1 1814	. . . 4 ⊢ (∀𝑥𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜑))
5	3, 4	syl6 35	. . 3 ⊢ (∀𝑥 𝑥 = 𝑡 → (∀𝑡𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜑)))
6	2, 5	syld 47	. 2 ⊢ (∀𝑥 𝑥 = 𝑡 → ([𝑡 / 𝑥]∀𝑡𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜑)))
7		sb4b 2499	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑡 → ([𝑡 / 𝑥]∀𝑡𝜑 ↔ ∀𝑥(𝑥 = 𝑡 → ∀𝑡𝜑)))
8		sp 2182	. . . . 5 ⊢ (∀𝑡𝜑 → 𝜑)
9	8	imim2i 16	. . . 4 ⊢ ((𝑥 = 𝑡 → ∀𝑡𝜑) → (𝑥 = 𝑡 → 𝜑))
10	9	alimi 1812	. . 3 ⊢ (∀𝑥(𝑥 = 𝑡 → ∀𝑡𝜑) → ∀𝑥(𝑥 = 𝑡 → 𝜑))
11	7, 10	syl6bi 255	. 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))

Syntax hints:  ¬ wn 3   → wi 4  ∀wal 1535  [wsb 2069

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-10 2145  ax-12 2177  ax-13 2390

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1781  df-nf 1785  df-sb 2070

This theorem is referenced by:  hbsb2a  2523  sb6f  2537