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 ( ( Ⅎ' 𝑥 𝜑 ∧ Ⅎ' 𝑥 𝜓 ) → Ⅎ' 𝑥 ( 𝜑 ↔ 𝜓 ) )

Proof

Step Hyp Ref Expression
1 bj-nnfim ⊢ ( ( Ⅎ' 𝑥 𝜑 ∧ Ⅎ' 𝑥 𝜓 ) → Ⅎ' 𝑥 ( 𝜑 → 𝜓 ) )
2 bj-nnfim ⊢ ( ( Ⅎ' 𝑥 𝜓 ∧ Ⅎ' 𝑥 𝜑 ) → Ⅎ' 𝑥 ( 𝜓 → 𝜑 ) )
3 2 ancoms ⊢ ( ( Ⅎ' 𝑥 𝜑 ∧ Ⅎ' 𝑥 𝜓 ) → Ⅎ' 𝑥 ( 𝜓 → 𝜑 ) )
4 bj-nnfan ⊢ ( ( Ⅎ' 𝑥 ( 𝜑 → 𝜓 ) ∧ Ⅎ' 𝑥 ( 𝜓 → 𝜑 ) ) → Ⅎ' 𝑥 ( ( 𝜑 → 𝜓 ) ∧ ( 𝜓 → 𝜑 ) ) )
5 1 3 4 syl2anc ⊢ ( ( Ⅎ' 𝑥 𝜑 ∧ Ⅎ' 𝑥 𝜓 ) → Ⅎ' 𝑥 ( ( 𝜑 → 𝜓 ) ∧ ( 𝜓 → 𝜑 ) ) )
6 dfbi2 ⊢ ( ( 𝜑 ↔ 𝜓 ) ↔ ( ( 𝜑 → 𝜓 ) ∧ ( 𝜓 → 𝜑 ) ) )
7 6 bicomi ⊢ ( ( ( 𝜑 → 𝜓 ) ∧ ( 𝜓 → 𝜑 ) ) ↔ ( 𝜑 ↔ 𝜓 ) )
8 7 bj-nnfbii ⊢ ( Ⅎ' 𝑥 ( ( 𝜑 → 𝜓 ) ∧ ( 𝜓 → 𝜑 ) ) ↔ Ⅎ' 𝑥 ( 𝜑 ↔ 𝜓 ) )
9 5 8 sylib ⊢ ( ( Ⅎ' 𝑥 𝜑 ∧ Ⅎ' 𝑥 𝜓 ) → Ⅎ' 𝑥 ( 𝜑 ↔ 𝜓 ) )