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Proof step 2 from proof gcd-rec-sub (previous step)

0 < i, i < j ⊦ gcd(i, j) = gcd(i, j - i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
1 < j, 0 < i, i < j, 0 < j ⊦ gcd(i, j) = gcd(i, j - i)

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs