Description: Nonfreeness in both sides implies nonfreeness in the biconditional, deduction form. (Contributed by BJ, 2-Dec-2023) (Proof modification is discouraged.)
Ref | Expression | ||
---|---|---|---|
Hypotheses | bj-nnfbid.1 | ⊢ ( 𝜑 → Ⅎ' 𝑥 𝜓 ) | |
bj-nnfbid.2 | ⊢ ( 𝜑 → Ⅎ' 𝑥 𝜒 ) | ||
Assertion | bj-nnfbid | ⊢ ( 𝜑 → Ⅎ' 𝑥 ( 𝜓 ↔ 𝜒 ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | bj-nnfbid.1 | ⊢ ( 𝜑 → Ⅎ' 𝑥 𝜓 ) | |
2 | bj-nnfbid.2 | ⊢ ( 𝜑 → Ⅎ' 𝑥 𝜒 ) | |
3 | bj-nnfim | ⊢ ( ( Ⅎ' 𝑥 𝜓 ∧ Ⅎ' 𝑥 𝜒 ) → Ⅎ' 𝑥 ( 𝜓 → 𝜒 ) ) | |
4 | 1 2 3 | syl2anc | ⊢ ( 𝜑 → Ⅎ' 𝑥 ( 𝜓 → 𝜒 ) ) |
5 | bj-nnfim | ⊢ ( ( Ⅎ' 𝑥 𝜒 ∧ Ⅎ' 𝑥 𝜓 ) → Ⅎ' 𝑥 ( 𝜒 → 𝜓 ) ) | |
6 | 2 1 5 | syl2anc | ⊢ ( 𝜑 → Ⅎ' 𝑥 ( 𝜒 → 𝜓 ) ) |
7 | 4 6 | bj-nnfand | ⊢ ( 𝜑 → Ⅎ' 𝑥 ( ( 𝜓 → 𝜒 ) ∧ ( 𝜒 → 𝜓 ) ) ) |
8 | dfbi2 | ⊢ ( ( 𝜓 ↔ 𝜒 ) ↔ ( ( 𝜓 → 𝜒 ) ∧ ( 𝜒 → 𝜓 ) ) ) | |
9 | 8 | bj-nnfbii | ⊢ ( Ⅎ' 𝑥 ( 𝜓 ↔ 𝜒 ) ↔ Ⅎ' 𝑥 ( ( 𝜓 → 𝜒 ) ∧ ( 𝜒 → 𝜓 ) ) ) |
10 | 7 9 | sylibr | ⊢ ( 𝜑 → Ⅎ' 𝑥 ( 𝜓 ↔ 𝜒 ) ) |