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Proof step 3 from proof l-num-03 (previous step)

n < n0, m0 + n0 ≤ n ⊦ n0 = 0

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: n0

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n;%nat freely generated by 0, +1;;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%nls-fst: ⊦ ¬ m < n → ¬ m + m0 < n; from specification nat-basic2