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Proof step 2 from proof succ-03 (previous step)

2^ m = 2 * n + 1 ⊦ m = 0 ∧ n = 0

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: m

Additional information

Used simplifier rules: Pot2-zero: ⊦ 2^ 0 = 1;%lemma-ax2: ⊦ 2^(n + 1) = 2 * 2^ 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% ⊦ 2 * m ≠ 2 * n + 1; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ 1 = m + 1 ↔ m = 0; (s)% ⊦ 2 * m = 0 ↔ m = 0; (s)