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Proof step 1 from proof exdivmod (conclusion)
j ≠ 0 ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) < abs(j)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
⊦ j = 0, ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) < abs(j)
Rule argument
Additional information