Theorem sbequ2 2285

Description: An equality theorem for substitution. (Contributed by NM, 16-May-1993.) Revise df-sb 2100. (Revised by BJ, 22-Dec-2020.) (Proof shortened by Wolf Lammen, 3-Feb-2024.)

Assertion

Ref	Expression
sbequ2	⊢ (𝑥 = 𝑡 → ([𝑡 / 𝑥]𝜑 → 𝜑))

Proof of Theorem sbequ2

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfsb 2101	. . . 4 ⊢ ([𝑡 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
2	1	biimpi 219	. . 3 ⊢ ([𝑡 / 𝑥]𝜑 → ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
3		equvinva 2063	. . 3 ⊢ (𝑥 = 𝑡 → ∃𝑦(𝑥 = 𝑦 ∧ 𝑡 = 𝑦))
4		equcomi 2050	. . . . . 6 ⊢ (𝑡 = 𝑦 → 𝑦 = 𝑡)
5		sp 2220	. . . . . 6 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 → 𝜑))
6	4, 5	imim12i 63	. . . . 5 ⊢ ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) → (𝑡 = 𝑦 → (𝑥 = 𝑦 → 𝜑)))
7	6	impcomd 417	. . . 4 ⊢ ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) → ((𝑥 = 𝑦 ∧ 𝑡 = 𝑦) → 𝜑))
8	7	aleximi 1865	. . 3 ⊢ (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) → (∃𝑦(𝑥 = 𝑦 ∧ 𝑡 = 𝑦) → ∃𝑦𝜑))
9	2, 3, 8	syl2im 41	. 2 ⊢ ([𝑡 / 𝑥]𝜑 → (𝑥 = 𝑡 → ∃𝑦𝜑))
10		ax5e 1945	. 2 ⊢ (∃𝑦𝜑 → 𝜑)
11	9, 10	syl6com 38	1 ⊢ (𝑥 = 𝑡 → ([𝑡 / 𝑥]𝜑 → 𝜑))

This proof depends on syntax axioms:   → 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-12 2213

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

This theorem is used by:  stdpc7  2286  sbequ12  2287  sb4a  2510  dfsb1  2511  dfsb2  2523  bj-ssbid2  37541  2pm13.193  45520  2pm13.193VD  45870