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Proof step 13 from proof gcd-one (previous step)

⊦ gcd(i, j) = 1, 1 < gcd(i, j), ∃ i0. 1 < i0 ∧ i0 | i ∧ i0 | j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
⊦ ∃ i. 1 < i

Rule argument

Additional information

Used simplifier rules: gcd-less-eq-zero: ⊦ gcd(i, j) ≤ 0 ↔ gcd(i, j) = 0;%gcd-zero-zero: ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0;%gcd-zero: ⊦ gcd(0, 0) = 0;%div-zero-01: ⊦ i | 0;% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < 1 ↔ i ≤ 0; (s)% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)