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Proof step 1 from proof zero-not (conclusion)
⊦ abs(i) < j ∧ j ≤ abs(i0) → i + i0 ≠ 0
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
i + i0 = 0, abs(i) < j ⊦ abs(i0) < j
Rule argument
Additional information
Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;