dfsb7

Description: An alternate definition of proper substitution df-sb . By introducing a dummy variable y in the definiens, we are able to eliminate any distinct variable restrictions among the variables t , x , and ph 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 , 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 . Theorem sb7h provides a version where ph and y don't have to be distinct. (Contributed by NM, 28-Jan-2004) Revise df-sb . (Revised by BJ, 25-Dec-2020) (Proof shortened by Wolf Lammen, 3-Sep-2023)

		Ref	Expression
	Assertion	dfsb7	$⊢ [t/ x] φ \leftrightarrow \exists y (y = t \land \exists x (x = y \land φ))$

Proof

Step	Hyp	Ref	Expression
1		sb56	$⊢ \exists y (y = t \land \forall x (x = y \to φ)) \leftrightarrow \forall y (y = t \to \forall x (x = y \to φ))$
2		sb56	$⊢ \exists x (x = y \land φ) \leftrightarrow \forall x (x = y \to φ)$
3	2	anbi2i	$⊢ (y = t \land \exists x (x = y \land φ)) \leftrightarrow (y = t \land \forall x (x = y \to φ))$
4	3	exbii	$⊢ \exists y (y = t \land \exists x (x = y \land φ)) \leftrightarrow \exists y (y = t \land \forall x (x = y \to φ))$
5		df-sb	$⊢ [t/ x] φ \leftrightarrow \forall y (y = t \to \forall x (x = y \to φ))$
6	1 4 5	3bitr4ri	$⊢ [t/ x] φ \leftrightarrow \exists y (y = t \land \exists x (x = y \land φ))$

Metamath Proof Explorer

Theorem dfsb7

Proof