Theorem sbequ1 2240

Description: An equality theorem for substitution. (Contributed by NM, 16-May-1993.) Revise df-sb 2068. (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 2027	. . . . 5 ⊢ (𝑥 = 𝑡 → (𝑦 = 𝑡 → 𝑥 = 𝑦))
2		ax12v 2172	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
3	1, 2	syl6 35	. . . 4 ⊢ (𝑥 = 𝑡 → (𝑦 = 𝑡 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
4	3	com23 86	. . 3 ⊢ (𝑥 = 𝑡 → (𝜑 → (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
5	4	alrimdv 1932	. 2 ⊢ (𝑥 = 𝑡 → (𝜑 → ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
6		df-sb 2068	. 2 ⊢ ([𝑡 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
7	5, 6	syl6ibr 251	1 ⊢ (𝑥 = 𝑡 → (𝜑 → [𝑡 / 𝑥]𝜑))

Syntax hints:   → wi 4  ∀wal 1537  [wsb 2067

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-12 2171

This theorem depends on definitions:  df-bi 206  df-an 397  df-ex 1783  df-sb 2068

This theorem is referenced by:  sbequ12  2244  dfsb1  2485  dfsb2  2497  2eu6  2658  bj-ssbid1  34845  sb5ALT  42145  2pm13.193  42172  2pm13.193VD  42523  sb5ALTVD  42533