sban

Description: Conjunction inside and outside of a substitution are equivalent. Compare 19.26 . (Contributed by NM, 14-May-1993) (Proof shortened by Steven Nguyen, 13-Aug-2023)

		Ref	Expression
	Assertion	sban	$⊢ [y/ x] (φ \land ψ) \leftrightarrow ([y/ x] φ \land [y/ x] ψ)$

Proof

Step	Hyp	Ref	Expression
1		simpl	$⊢ (φ \land ψ) \to φ$
2	1	sbimi	$⊢ [y/ x] (φ \land ψ) \to [y/ x] φ$
3		simpr	$⊢ (φ \land ψ) \to ψ$
4	3	sbimi	$⊢ [y/ x] (φ \land ψ) \to [y/ x] ψ$
5	2 4	jca	$⊢ [y/ x] (φ \land ψ) \to ([y/ x] φ \land [y/ x] ψ)$
6		pm3.2	$⊢ φ \to (ψ \to (φ \land ψ))$
7	6	sb2imi	$⊢ [y/ x] φ \to ([y/ x] ψ \to [y/ x] (φ \land ψ))$
8	7	imp	$⊢ ([y/ x] φ \land [y/ x] ψ) \to [y/ x] (φ \land ψ)$
9	5 8	impbii	$⊢ [y/ x] (φ \land ψ) \leftrightarrow ([y/ x] φ \land [y/ x] ψ)$

Metamath Proof Explorer

Theorem sban

Proof