Theorem bj-dfssb2 32640

Description: An alternate definition of df-ssb 32620. Note that the use of a dummy variable in the definition df-ssb 32620 allows to use bj-sb56 32639 instead of equs45f 2350 and hence to avoid dependency on ax-13 2246 and to use ax-12 2047 only through bj-ax12 32634. Compare dfsb7 2455. (Contributed by BJ, 25-Dec-2020.)

Assertion

Ref	Expression
bj-dfssb2	⊢ ([𝑡/𝑥]b𝜑 ↔ ∃𝑦(𝑦 = 𝑡 ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))

Distinct variable groups:   𝑦,𝑡   𝑥,𝑦   𝜑,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑡)

Proof of Theorem bj-dfssb2

Step	Hyp	Ref	Expression
1		df-ssb 32620	. 2 ⊢ ([𝑡/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
2		bj-sb56 32639	. 2 ⊢ (∃𝑦(𝑦 = 𝑡 ∧ ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
3		bj-sb56 32639	. . . . 5 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
4	3	bicomi 214	. . . 4 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
5	4	anbi2i 730	. . 3 ⊢ ((𝑦 = 𝑡 ∧ ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ (𝑦 = 𝑡 ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
6	5	exbii 1774	. 2 ⊢ (∃𝑦(𝑦 = 𝑡 ∧ ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ ∃𝑦(𝑦 = 𝑡 ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
7	1, 2, 6	3bitr2i 288	1 ⊢ ([𝑡/𝑥]b𝜑 ↔ ∃𝑦(𝑦 = 𝑡 ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))

Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 384  ∀wal 1481  ∃wex 1704  [wssb 32619

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-an 386  df-ex 1705  df-ssb 32620

This theorem is referenced by:  bj-ssbn  32641