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Proof step 2 from proof num-i2n-le-03 (previous step)

i < 0 ⊦ m ≤ i→n(i) ↔ n→i(m) ≤ ~ i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ ~ j ↔ i + j ≤ 0; (s)