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Proof step 2 from proof num-i2n-ls-01 (previous step)

i < 0 ⊦ i→n(i) < m ↔ ~ i < n→i(m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i < 0 ⊦ i→n(i) < m ↔ 0 < n→i(m) + i

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ ~ i < j ↔ 0 < j + i; (s)