Theorem bj-dfsb2 36839

Description: Alternate (dual) definition of substitution df-sb 2065 not using dummy variables. (Contributed by BJ, 19-Mar-2021.)

Assertion

Ref	Expression
bj-dfsb2	⊢ ([𝑦 / 𝑥]𝜑 ↔ (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑)))

Proof of Theorem bj-dfsb2

Step	Hyp	Ref	Expression
1		dfsb1 2486	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
2		bj-sbsb 36838	. 2 ⊢ (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ↔ (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑)))
3	1, 2	bitri 275	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   ∨ wo 848  ∀wal 1538  ∃wex 1779  [wsb 2064

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-10 2141  ax-12 2177  ax-13 2377

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-ex 1780  df-nf 1784  df-sb 2065

This theorem is referenced by: (None)