Theorem sbequ1 2287

Description: An equality theorem for substitution. (Contributed by NM, 16-May-1993.) Revise df-sb 2100. (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 2057	. . . . 5 ⊢ (𝑥 = 𝑡 → (𝑦 = 𝑡 → 𝑥 = 𝑦))
2		ax12v 2217	. . . . 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 1962	. 2 ⊢ (𝑥 = 𝑡 → (𝜑 → ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
6		dfsb 2101	. 2 ⊢ ([𝑡 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
7	5, 6	imbitrrdi 255	1 ⊢ (𝑥 = 𝑡 → (𝜑 → [𝑡 / 𝑥]𝜑))

This proof depends on syntax axioms:   → 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-12 2216

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

This theorem is used by:  sbequ12  2290  dfsb1  2516  dfsb2  2528  2eu6  2687  bj-ssbid1  37327  sb5ALT  45275  2pm13.193  45302  2pm13.193VD  45652  sb5ALTVD  45662