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Description: For each finite subset of the integers not containing 0 there is a positive integer which is divisible by each element of this subset. (Contributed by AV, 21-Aug-2020)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | fissn0dvds | ⊢ ( ( 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍 ) → ∃ 𝑛 ∈ ℕ ∀ 𝑚 ∈ 𝑍 𝑚 ∥ 𝑛 ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | simp1 | ⊢ ( ( 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍 ) → 𝑍 ⊆ ℤ ) | |
| 2 | simp2 | ⊢ ( ( 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍 ) → 𝑍 ∈ Fin ) | |
| 3 | eqid | ⊢ ( abs ‘ ∏ 𝑘 ∈ 𝑍 𝑘 ) = ( abs ‘ ∏ 𝑘 ∈ 𝑍 𝑘 ) | |
| 4 | simp3 | ⊢ ( ( 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍 ) → 0 ∉ 𝑍 ) | |
| 5 | 1 2 3 4 | absprodnn | ⊢ ( ( 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍 ) → ( abs ‘ ∏ 𝑘 ∈ 𝑍 𝑘 ) ∈ ℕ ) |
| 6 | breq2 | ⊢ ( 𝑛 = ( abs ‘ ∏ 𝑘 ∈ 𝑍 𝑘 ) → ( 𝑚 ∥ 𝑛 ↔ 𝑚 ∥ ( abs ‘ ∏ 𝑘 ∈ 𝑍 𝑘 ) ) ) | |
| 7 | 6 | ralbidv | ⊢ ( 𝑛 = ( abs ‘ ∏ 𝑘 ∈ 𝑍 𝑘 ) → ( ∀ 𝑚 ∈ 𝑍 𝑚 ∥ 𝑛 ↔ ∀ 𝑚 ∈ 𝑍 𝑚 ∥ ( abs ‘ ∏ 𝑘 ∈ 𝑍 𝑘 ) ) ) |
| 8 | 7 | adantl | ⊢ ( ( ( 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍 ) ∧ 𝑛 = ( abs ‘ ∏ 𝑘 ∈ 𝑍 𝑘 ) ) → ( ∀ 𝑚 ∈ 𝑍 𝑚 ∥ 𝑛 ↔ ∀ 𝑚 ∈ 𝑍 𝑚 ∥ ( abs ‘ ∏ 𝑘 ∈ 𝑍 𝑘 ) ) ) |
| 9 | 1 2 3 | absproddvds | ⊢ ( ( 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍 ) → ∀ 𝑚 ∈ 𝑍 𝑚 ∥ ( abs ‘ ∏ 𝑘 ∈ 𝑍 𝑘 ) ) |
| 10 | 5 8 9 | rspcedvd | ⊢ ( ( 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍 ) → ∃ 𝑛 ∈ ℕ ∀ 𝑚 ∈ 𝑍 𝑚 ∥ 𝑛 ) |