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Description: Deduction eliminating an inequality in an antecedent. (Contributed by NM, 24-May-2006) (Proof shortened by Andrew Salmon, 25-May-2011) (Proof shortened by Wolf Lammen, 25-Nov-2019)
| Ref | Expression | ||
|---|---|---|---|
| Hypotheses | pm2.61ne.1 | ⊢ ( 𝐴 = 𝐵 → ( 𝜓 ↔ 𝜒 ) ) | |
| pm2.61ne.2 | ⊢ ( ( 𝜑 ∧ 𝐴 ≠ 𝐵 ) → 𝜓 ) | ||
| pm2.61ne.3 | ⊢ ( 𝜑 → 𝜒 ) | ||
| Assertion | pm2.61ne | ⊢ ( 𝜑 → 𝜓 ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | pm2.61ne.1 | ⊢ ( 𝐴 = 𝐵 → ( 𝜓 ↔ 𝜒 ) ) | |
| 2 | pm2.61ne.2 | ⊢ ( ( 𝜑 ∧ 𝐴 ≠ 𝐵 ) → 𝜓 ) | |
| 3 | pm2.61ne.3 | ⊢ ( 𝜑 → 𝜒 ) | |
| 4 | 3 1 | imbitrrid | ⊢ ( 𝐴 = 𝐵 → ( 𝜑 → 𝜓 ) ) |
| 5 | 2 | expcom | ⊢ ( 𝐴 ≠ 𝐵 → ( 𝜑 → 𝜓 ) ) |
| 6 | 4 5 | pm2.61ine | ⊢ ( 𝜑 → 𝜓 ) |