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Proof step 2 from proof le-one (previous step)

n < n0, m ≤ n + 1 ⊦ m ≤ n0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ m ≤ n ↔ ¬ n < m;%fle: ⊦ m < n ∧ n ≤ n0 → m < n0;%succ-02: ⊦ m < n + 1 ↔ m ≤ n;%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic