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Proof step 2 from proof less-03 (previous step)

m - n0 < m - n1, n0 ≤ m, n1 ≤ m ⊦ n1 < n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m < (m - n1) + n0, n0 ≤ n1, n0 ≤ m, n1 ≤ m ⊦ m + n0 = n1

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ m ≤ n ↔ ¬ n < m;%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false);%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0;%lels: ⊦ m < n ↔ ¬ n ≤ m;%eq-02: n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic