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Proof step 9 from proof two-loup-12 (previous step)

m + n = m0 * 2 + 1 ⊦ m0 < n ↔ m < n

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%twoeqf-03: ⊦ m + m0 = 2 * n + 1 ∧ m < n → n + 1 < m0; (forward) from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%twoeqf-02: ⊦ m + m0 = 2 * n + 1 ∧ n < m → m0 ≤ n; (forward) from specification nat-mult%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat