Theorem sbequ1 2286

Description: An equality theorem for substitution. (Contributed by NM, 16-May-1993.) Revise df-sb 2094. (Revised by BJ, 22-Dec-2020.)

Assertion

Ref	Expression
sbequ1	⊢ (𝑥 = 𝑡 → (𝜑 → [𝑡 / 𝑥]𝜑))

Proof of Theorem sbequ1

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		equeucl 2047	. . . . 5 ⊢ (𝑥 = 𝑡 → (𝑦 = 𝑡 → 𝑥 = 𝑦))
2		ax12v 2216	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
3	1, 2	syl6 36	. . . 4 ⊢ (𝑥 = 𝑡 → (𝑦 = 𝑡 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
4	3	com23 87	. . 3 ⊢ (𝑥 = 𝑡 → (𝜑 → (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
5	4	alrimdv 1952	. 2 ⊢ (𝑥 = 𝑡 → (𝜑 → ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
6		dfsb 2096	. 2 ⊢ ([𝑡 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
7	5, 6	imbitrrdi 255	1 ⊢ (𝑥 = 𝑡 → (𝜑 → [𝑡 / 𝑥]𝜑))

Syntax hints:   → wi 4  ∀wal 1561  [wsb 2093

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-12 2215

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1803  df-sb 2094

This theorem is referenced by:  sbequ12  2289  dfsb1  2515  dfsb2  2527  2eu6  2686  bj-ssbid1  37148  sb5ALT  45099  2pm13.193  45126  2pm13.193VD  45476  sb5ALTVD  45486