Theorem bj-sb6 32767

Description: Remove dependency on ax-13 2246 from sb6 2429. (Contributed by BJ, 11-Sep-2019.) (Proof modification is discouraged.)

Assertion

Ref	Expression
bj-sb6	⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem bj-sb6

Step	Hyp	Ref	Expression
1		sb1 1883	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
2		sb56 2150	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
3	1, 2	sylib 208	. 2 ⊢ ([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))
4		bj-sb2v 32753	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜑)
5	3, 4	impbii 199	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 384  ∀wal 1481  ∃wex 1704  [wsb 1880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-12 2047

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1705  df-nf 1710  df-sb 1881

This theorem is referenced by:  bj-sb5  32768