Description: Introduce a right anti-conjunct to both sides of a logical equivalence. (Contributed by Anthony Hart, 1-Sep-2011) (Proof shortened by Wolf Lammen, 27-Jun-2020)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | nanbi1 | ⊢ ( ( 𝜑 ↔ 𝜓 ) → ( ( 𝜑 ⊼ 𝜒 ) ↔ ( 𝜓 ⊼ 𝜒 ) ) ) | 
| Step | Hyp | Ref | Expression | 
|---|---|---|---|
| 1 | imbi1 | ⊢ ( ( 𝜑 ↔ 𝜓 ) → ( ( 𝜑 → ¬ 𝜒 ) ↔ ( 𝜓 → ¬ 𝜒 ) ) ) | |
| 2 | dfnan2 | ⊢ ( ( 𝜑 ⊼ 𝜒 ) ↔ ( 𝜑 → ¬ 𝜒 ) ) | |
| 3 | dfnan2 | ⊢ ( ( 𝜓 ⊼ 𝜒 ) ↔ ( 𝜓 → ¬ 𝜒 ) ) | |
| 4 | 1 2 3 | 3bitr4g | ⊢ ( ( 𝜑 ↔ 𝜓 ) → ( ( 𝜑 ⊼ 𝜒 ) ↔ ( 𝜓 ⊼ 𝜒 ) ) ) |