bj-nnfan

Description: Nonfreeness in both conjuncts implies nonfreeness in the conjunction. (Contributed by BJ, 19-Nov-2023) In classical logic, there is a proof using the definition of conjunction in terms of implication and negation, so using bj-nnfim , bj-nnfnt and bj-nnfbi , but we want a proof valid in intuitionistic logic. (Proof modification is discouraged.)

		Ref	Expression
	Assertion	bj-nnfan	$⊢ (Ⅎ' x φ \land Ⅎ' x ψ) \to Ⅎ' x (φ \land ψ)$

Proof

Step	Hyp	Ref	Expression
1		df-bj-nnf	$⊢ Ⅎ' x φ \leftrightarrow ((\exists x φ \to φ) \land (φ \to \forall x φ))$
2		df-bj-nnf	$⊢ Ⅎ' x ψ \leftrightarrow ((\exists x ψ \to ψ) \land (ψ \to \forall x ψ))$
3		19.40	$⊢ \exists x (φ \land ψ) \to (\exists x φ \land \exists x ψ)$
4		anim12	$⊢ ((\exists x φ \to φ) \land (\exists x ψ \to ψ)) \to ((\exists x φ \land \exists x ψ) \to (φ \land ψ))$
5	3 4	syl5	$⊢ ((\exists x φ \to φ) \land (\exists x ψ \to ψ)) \to (\exists x (φ \land ψ) \to (φ \land ψ))$
6		anim12	$⊢ ((φ \to \forall x φ) \land (ψ \to \forall x ψ)) \to ((φ \land ψ) \to (\forall x φ \land \forall x ψ))$
7		id	$⊢ (φ \land ψ) \to (φ \land ψ)$
8	7	alanimi	$⊢ (\forall x φ \land \forall x ψ) \to \forall x (φ \land ψ)$
9	6 8	syl6	$⊢ ((φ \to \forall x φ) \land (ψ \to \forall x ψ)) \to ((φ \land ψ) \to \forall x (φ \land ψ))$
10	5 9	anim12i	$⊢ (((\exists x φ \to φ) \land (\exists x ψ \to ψ)) \land ((φ \to \forall x φ) \land (ψ \to \forall x ψ))) \to ((\exists x (φ \land ψ) \to (φ \land ψ)) \land ((φ \land ψ) \to \forall x (φ \land ψ)))$
11	10	an4s	$⊢ (((\exists x φ \to φ) \land (φ \to \forall x φ)) \land ((\exists x ψ \to ψ) \land (ψ \to \forall x ψ))) \to ((\exists x (φ \land ψ) \to (φ \land ψ)) \land ((φ \land ψ) \to \forall x (φ \land ψ)))$
12	1 2 11	syl2anb	$⊢ (Ⅎ' x φ \land Ⅎ' x ψ) \to ((\exists x (φ \land ψ) \to (φ \land ψ)) \land ((φ \land ψ) \to \forall x (φ \land ψ)))$
13		df-bj-nnf	$⊢ Ⅎ' x (φ \land ψ) \leftrightarrow ((\exists x (φ \land ψ) \to (φ \land ψ)) \land ((φ \land ψ) \to \forall x (φ \land ψ)))$
14	12 13	sylibr	$⊢ (Ⅎ' x φ \land Ⅎ' x ψ) \to Ⅎ' x (φ \land ψ)$

Metamath Proof Explorer

Theorem bj-nnfan

Proof