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Proof step 2 from proof le-num-01 (previous step)

n < n0 ⊦ m + n ≤ m0 + n0 ↔ m ≤ m0 + (n0 - n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < n0 ⊦ m + n ≤ m0 + n0 ↔ m ≤ m0 + (n0 - n)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic