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

⊦ 1 < gcd(i, j), gcd(i, j) = 1, ∃ 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;%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1