Theorem bj-dfsb2 36882

Description: Alternate (dual) definition of substitution df-sb 2068 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 2481	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
2		bj-sbsb 36881	. 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 847  ∀wal 1539  ∃wex 1780  [wsb 2067

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-10 2144  ax-12 2180  ax-13 2372

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-ex 1781  df-nf 1785  df-sb 2068

This theorem is referenced by: (None)