Theorem bj-dfsb2 36804

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 2489	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
2		bj-sbsb 36803	. 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 846  ∀wal 1535  ∃wex 1777  [wsb 2064

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-10 2141  ax-12 2178  ax-13 2380

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-ex 1778  df-nf 1782  df-sb 2065

This theorem is referenced by: (None)