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Proof step 2 from proof two-succ-04 (previous step)

⊦ 2 * m + 1 < 2 * n ↔ m < n

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-02: ⊦ m * (n + 1) = m * n + m;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%two-ls-01: ⊦ 2 * m < 2 * n ↔ m < n;%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%n < n0 ⊦ m + (* n) < m0 + (* n0) ↔ m < m0 + (n0 - n); (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%n0 < n ⊦ ¬ m + (* n) < (* n0); (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic