Description: A contradiction concerning restricted generalization for a nonempty set implies anything. (Contributed by Alexander van der Vekens, 4-Sep-2018)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | ralnralall | ⊢ ( 𝐴 ≠ ∅ → ( ( ∀ 𝑥 ∈ 𝐴 𝜑 ∧ ∀ 𝑥 ∈ 𝐴 ¬ 𝜑 ) → 𝜓 ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | r19.26 | ⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ∧ ¬ 𝜑 ) ↔ ( ∀ 𝑥 ∈ 𝐴 𝜑 ∧ ∀ 𝑥 ∈ 𝐴 ¬ 𝜑 ) ) | |
| 2 | pm3.24 | ⊢ ¬ ( 𝜑 ∧ ¬ 𝜑 ) | |
| 3 | 2 | bifal | ⊢ ( ( 𝜑 ∧ ¬ 𝜑 ) ↔ ⊥ ) |
| 4 | 3 | ralbii | ⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ∧ ¬ 𝜑 ) ↔ ∀ 𝑥 ∈ 𝐴 ⊥ ) |
| 5 | r19.3rzv | ⊢ ( 𝐴 ≠ ∅ → ( ⊥ ↔ ∀ 𝑥 ∈ 𝐴 ⊥ ) ) | |
| 6 | falim | ⊢ ( ⊥ → 𝜓 ) | |
| 7 | 5 6 | syl6bir | ⊢ ( 𝐴 ≠ ∅ → ( ∀ 𝑥 ∈ 𝐴 ⊥ → 𝜓 ) ) |
| 8 | 4 7 | syl5bi | ⊢ ( 𝐴 ≠ ∅ → ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ∧ ¬ 𝜑 ) → 𝜓 ) ) |
| 9 | 1 8 | syl5bir | ⊢ ( 𝐴 ≠ ∅ → ( ( ∀ 𝑥 ∈ 𝐴 𝜑 ∧ ∀ 𝑥 ∈ 𝐴 ¬ 𝜑 ) → 𝜓 ) ) |