Theorem dfsb7 2316

Description: An alternate definition of proper substitution df-sb 2100. By introducing a dummy variable 𝑦 in the definiens, we are able to eliminate any distinct variable restrictions among the variables 𝑡, 𝑥, and 𝜑 of the definiendum. No distinct variable conflicts arise because 𝑦 effectively insulates 𝑡 from 𝑥. To achieve this, we use a chain of two substitutions in the form of sb5 2313, first 𝑦 for 𝑥 then 𝑡 for 𝑦. Compare Definition 2.1'' of [Quine] p. 17, which is obtained from this theorem by applying df-clab 2744. Theorem sb7h 2560 provides a version where 𝜑 and 𝑦 don't have to be distinct. (Contributed by NM, 28-Jan-2004.) Revise df-sb 2100. (Revised by BJ, 25-Dec-2020.) (Proof shortened by Wolf Lammen, 3-Sep-2023.)

Assertion

Ref	Expression
dfsb7	⊢ ([𝑡 / 𝑥]𝜑 ↔ ∃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 dfsb7

Step	Hyp	Ref	Expression
1		sbalex 2281	. 2 ⊢ (∃𝑦(𝑦 = 𝑡 ∧ ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
2		sbalex 2281	. . . 4 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
3	2	anbi2i 635	. . 3 ⊢ ((𝑦 = 𝑡 ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ↔ (𝑦 = 𝑡 ∧ ∀𝑥(𝑥 = 𝑦 → 𝜑)))
4	3	exbii 1881	. 2 ⊢ (∃𝑦(𝑦 = 𝑡 ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ↔ ∃𝑦(𝑦 = 𝑡 ∧ ∀𝑥(𝑥 = 𝑦 → 𝜑)))
5		dfsb 2101	. 2 ⊢ ([𝑡 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
6	1, 4, 5	3bitr4ri 307	1 ⊢ ([𝑡 / 𝑥]𝜑 ↔ ∃𝑦(𝑦 = 𝑡 ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812  [wsb 2099

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2179  ax-12 2216

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817  df-sb 2100

This theorem is used by:  sbn  2317