a51139e63b
Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
37 lines
1.5 KiB
Text
37 lines
1.5 KiB
Text
import macros
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import tactic
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using Nat
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definition dvd (a b : Nat) : Bool := ∃ c, a * c = b
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infix 50 | : dvd
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theorem dvd_elim {a b : Nat} (H : a | b) : ∃ c, a * c = b := H
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theorem dvd_intro {a b : Nat} (c : Nat) (H : a * c = b) : a | b := exists_intro c H
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set_opaque dvd true
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theorem dvd_trans {a b c} (H1 : a | b) (H2 : b | c) : a | c
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:= obtain (w1 : Nat) (Hw1 : a * w1 = b), from dvd_elim H1,
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obtain (w2 : Nat) (Hw2 : b * w2 = c), from dvd_elim H2,
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dvd_intro (w1 * w2)
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(calc a * (w1 * w2) = a * w1 * w2 : symm (mul_assoc a w1 w2)
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... = b * w2 : { Hw1 }
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... = c : Hw2)
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definition prime p := p ≥ 2 ∧ forall m, m | p → m = 1 ∨ m = p
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theorem not_prime_eq (n : Nat) (H1 : n ≥ 2) (H2 : ¬ prime n) : ∃ m, m | n ∧ m ≠ 1 ∧ m ≠ n
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:= have H3 : ¬ n ≥ 2 ∨ ¬ (∀ m : Nat, m | n → m = 1 ∨ m = n),
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from (not_and _ _ ◂ H2),
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have H4 : ¬ ¬ n ≥ 2,
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by skip, -- Ignore this hole
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obtain (m : Nat) (H5 : ¬ (m | n → m = 1 ∨ m = n)),
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from (not_forall_elim (resolve1 H3 H4)),
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have H6 : m | n ∧ ¬ (m = 1 ∨ m = n),
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by id, -- <<< id is the "do-nothing" tactic, it will fail and Lean will display the proof state for this hole
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have H7 : ¬ (m = 1 ∨ m = n) ↔ (m ≠ 1 ∧ m ≠ n),
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from (not_or (m = 1) (m = n)),
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have H8 : m | n ∧ m ≠ 1 ∧ m ≠ n,
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from subst H6 H7,
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show ∃ m, m | n ∧ m ≠ 1 ∧ m ≠ n,
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from exists_intro m H8
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