Theorem bj-nnfand

Description: Nonfreeness in both conjuncts implies nonfreeness in the conjunction, deduction form. Note: compared with the proof of bj-nnfan , it has two more essential steps but fewer total steps (since there are fewer intermediate formulas to build) and is easier to follow and understand. This statement is of intermediate complexity: for simpler statements, closed-style proofs like that of bj-nnfan will generally be shorter than deduction-style proofs while still easy to follow, while for more complex statements, the opposite will be true (and deduction-style proofs like that of bj-nnfand will generally be easier to understand). (Contributed by BJ, 19-Nov-2023) (Proof modification is discouraged.)

	Ref	Expression
Hypotheses	bj-nnfand.1	|- ( ph -> F// x ps )
	bj-nnfand.2	|- ( ph -> F// x ch )
Assertion	bj-nnfand	|- ( ph -> F// x ( ps /\ ch ) )

Proof

Step	Hyp	Ref	Expression
1		bj-nnfand.1	|- ( ph -> F// x ps )
2		bj-nnfand.2	|- ( ph -> F// x ch )
3		19.40	|- ( E. x ( ps /\ ch ) -> ( E. x ps /\ E. x ch ) )
4	1	bj-nnfed	|- ( ph -> ( E. x ps -> ps ) )
5	2	bj-nnfed	|- ( ph -> ( E. x ch -> ch ) )
6	4 5	anim12d	|- ( ph -> ( ( E. x ps /\ E. x ch ) -> ( ps /\ ch ) ) )
7	3 6	syl5	|- ( ph -> ( E. x ( ps /\ ch ) -> ( ps /\ ch ) ) )
8	1	bj-nnfad	|- ( ph -> ( ps -> A. x ps ) )
9	2	bj-nnfad	|- ( ph -> ( ch -> A. x ch ) )
10	8 9	anim12d	|- ( ph -> ( ( ps /\ ch ) -> ( A. x ps /\ A. x ch ) ) )
11		19.26	|- ( A. x ( ps /\ ch ) <-> ( A. x ps /\ A. x ch ) )
12	10 11	syl6ibr	|- ( ph -> ( ( ps /\ ch ) -> A. x ( ps /\ ch ) ) )
13		df-bj-nnf	|- ( F// x ( ps /\ ch ) <-> ( ( E. x ( ps /\ ch ) -> ( ps /\ ch ) ) /\ ( ( ps /\ ch ) -> A. x ( ps /\ ch ) ) ) )
14	7 12 13	sylanbrc	|- ( ph -> F// x ( ps /\ ch ) )