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Metamath Proof Explorer

Theorem bj-nnfbit

Description: Nonfreeness in both sides implies nonfreeness in the biconditional. (Contributed by BJ, 2-Dec-2023) (Proof modification is discouraged.)

Ref Expression
Assertion bj-nnfbit 
|- ( ( F// x ph /\ F// x ps ) -> F// x ( ph <-> ps ) )

Proof

Step Hyp Ref Expression
1 bj-nnfim 
 |-  ( ( F// x ph /\ F// x ps ) -> F// x ( ph -> ps ) )
2 bj-nnfim 
 |-  ( ( F// x ps /\ F// x ph ) -> F// x ( ps -> ph ) )
3 2 ancoms 
 |-  ( ( F// x ph /\ F// x ps ) -> F// x ( ps -> ph ) )
4 bj-nnfan 
 |-  ( ( F// x ( ph -> ps ) /\ F// x ( ps -> ph ) ) -> F// x ( ( ph -> ps ) /\ ( ps -> ph ) ) )
5 1 3 4 syl2anc 
 |-  ( ( F// x ph /\ F// x ps ) -> F// x ( ( ph -> ps ) /\ ( ps -> ph ) ) )
6 dfbi2 
 |-  ( ( ph <-> ps ) <-> ( ( ph -> ps ) /\ ( ps -> ph ) ) )
7 6 bicomi 
 |-  ( ( ( ph -> ps ) /\ ( ps -> ph ) ) <-> ( ph <-> ps ) )
8 7 bj-nnfbii 
 |-  ( F// x ( ( ph -> ps ) /\ ( ps -> ph ) ) <-> F// x ( ph <-> ps ) )
9 5 8 sylib 
 |-  ( ( F// x ph /\ F// x ps ) -> F// x ( ph <-> ps ) )