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Proof step 4 from proof le-add-01 (previous step)

n < m, ∃ m1. n0 + m1 = m0 ∧ m1 ≠ 0, m + m0 ≤ n + n0 ⊦ m = 0, m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n < m, m + m1 ≤ n ⊦ m = 0, n0 + m1 = 0, m1 = 0

Rule argument

Additional information

Used simplifier rules: lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2