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Proof step 2 from proof threels (previous step)

⊦ 3 < 2 * n ↔ n ≠ 0 ∧ n ≠ 1

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: n

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%rsucc-01: ⊦ (m + 1) * n = m * n + n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%ls2-03: ⊦ 1 < 2 * n ↔ n ≠ 0;%nat freely generated by 0, +1;;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%l-num-04: n0 < n ⊦ (* n) < m0 + (* n0) ↔ n - n0 < m0; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%one-03: ⊦ 1 = m + 1 ↔ m = 0; from specification nat-basic2