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Proof step 3 from proof le-num-06 (previous step)

n0 < n, m + n ≤ n0 ⊦

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: n

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n;%succ-02: ⊦ m < n + 1 ↔ m ≤ n;%fle: ⊦ m < n ∧ n ≤ n0 → m < n0;%nat freely generated by 0, +1;;% ⊦ m +1 = m + 1; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ m + n < m; (s)% ⊦ ¬ m < 0; (s)