Theorem dfsb 2100

Description: Simplify definition df-sb 2099 by removing its provable hypothesis. (Contributed by Wolf Lammen, 5-Feb-2026.)

Assertion

Ref	Expression
dfsb	⊢ ([𝑡 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))

Distinct variable groups:   𝑥,𝑦   𝑦,𝑡   𝜑,𝑦

Allowed substitution hints:   𝜑(𝑥, 𝑡)

Proof of Theorem dfsb

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbjust 2097	. 2 ⊢ (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ ∀𝑧(𝑧 = 𝑡 → ∀𝑥(𝑥 = 𝑧 → 𝜑)))
2	1	df-sb 2099	1 ⊢ ([𝑡 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567  [wsb 2098

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-sb 2099

This theorem is used by:  stdpc4  2104  sbi1  2107  spsbe  2118  sbequ  2119  sb6  2121  sbal  2205  sbequ1  2283  sbequ2  2284  dfsb7  2312  sbn  2313  sbrim  2337  cbvsbvf  2392  sb4b  2504  sbequbidv  37095  cbvsbdavw  37135  cbvsbdavw2  37136  bj-ssbeq  37644  bj-ssbid2ALT  37654  bj-ssbid1ALT  37656