Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
||
Mirrors > Home > MPE Home > Th. List > nanbi1 | Structured version Visualization version GIF version |
Description: Introduce a right anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.) |
Ref | Expression |
---|---|
nanbi1 | ⊢ ((𝜑 ↔ 𝜓) → ((𝜑 ⊼ 𝜒) ↔ (𝜓 ⊼ 𝜒))) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | anbi1 743 | . . 3 ⊢ ((𝜑 ↔ 𝜓) → ((𝜑 ∧ 𝜒) ↔ (𝜓 ∧ 𝜒))) | |
2 | 1 | notbid 308 | . 2 ⊢ ((𝜑 ↔ 𝜓) → (¬ (𝜑 ∧ 𝜒) ↔ ¬ (𝜓 ∧ 𝜒))) |
3 | df-nan 1448 | . 2 ⊢ ((𝜑 ⊼ 𝜒) ↔ ¬ (𝜑 ∧ 𝜒)) | |
4 | df-nan 1448 | . 2 ⊢ ((𝜓 ⊼ 𝜒) ↔ ¬ (𝜓 ∧ 𝜒)) | |
5 | 2, 3, 4 | 3bitr4g 303 | 1 ⊢ ((𝜑 ↔ 𝜓) → ((𝜑 ⊼ 𝜒) ↔ (𝜓 ⊼ 𝜒))) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 → wi 4 ↔ wb 196 ∧ wa 384 ⊼ wnan 1447 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 |
This theorem depends on definitions: df-bi 197 df-an 386 df-nan 1448 |
This theorem is referenced by: nanbi2 1456 nanbi12 1457 nanbi1i 1458 nanbi1d 1461 nabi1 32389 |
Copyright terms: Public domain | W3C validator |