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Description: Eliminate an antecedent implied by each side of a biconditional. Variant of pm5.21ndd . (Contributed by NM, 20-Nov-2005) (Proof shortened by Wolf Lammen, 4-Nov-2013)
| Ref | Expression | ||
|---|---|---|---|
| Hypotheses | pm5.21nd.1 | ⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝜃 ) | |
| pm5.21nd.2 | ⊢ ( ( 𝜑 ∧ 𝜒 ) → 𝜃 ) | ||
| pm5.21nd.3 | ⊢ ( 𝜃 → ( 𝜓 ↔ 𝜒 ) ) | ||
| Assertion | pm5.21nd | ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | pm5.21nd.1 | ⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝜃 ) | |
| 2 | pm5.21nd.2 | ⊢ ( ( 𝜑 ∧ 𝜒 ) → 𝜃 ) | |
| 3 | pm5.21nd.3 | ⊢ ( 𝜃 → ( 𝜓 ↔ 𝜒 ) ) | |
| 4 | 1 | ex | ⊢ ( 𝜑 → ( 𝜓 → 𝜃 ) ) |
| 5 | 2 | ex | ⊢ ( 𝜑 → ( 𝜒 → 𝜃 ) ) |
| 6 | 3 | a1i | ⊢ ( 𝜑 → ( 𝜃 → ( 𝜓 ↔ 𝜒 ) ) ) |
| 7 | 4 5 6 | pm5.21ndd | ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) ) |