Theorem bj-dfsb2 34948

Description: Alternate (dual) definition of substitution df-sb 2069 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 2485	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
2		bj-sbsb 34947	. 2 ⊢ (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ↔ (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑)))
3	1, 2	bitri 274	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   ∨ wo 843  ∀wal 1537  ∃wex 1783  [wsb 2068

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-10 2139  ax-12 2173  ax-13 2372

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-ex 1784  df-nf 1788  df-sb 2069

This theorem is referenced by: (None)