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Proof step 2 from proof le2-04 (previous step)

⊦ 2 * n ≤ 2 * m + 1 ↔ n ≤ m

User applied rule case distinction

And obtained the following new goals

Proof step 3:
2 * n ≤ 2 * m + 1 ⊦ n ≤ m

Rule argument

pos: right 1

Additional information

Used simplifier rules: modls: ⊦ n0 < m0 → (m0 * m + n0 < m0 * n ↔ m < n);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2