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Proof step 1 from proof gcd-prime (conclusion)

⊦ prime(j) ∧ 1 < i ∧ i < j → gcd(i, j) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
1 < i, 2 < j, 0 < i, i < j, prime(j), 0 < j, 1 < j ⊦ gcd(i, j) = 1

Rule argument

Additional information

Used simplifier rules: numlslsdel-01: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j; from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1