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Proof step 2 from proof pos-add (previous step)

n→i(m) + n→i(n) < j ⊦ 0 < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)