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Proof step 1 from proof pot2-def (conclusion)

⊦ 2^ m = 2 ^ m

User applied rule structural induction

And closed the goal!

Rule argument

indvar: m

Additional information

Used simplifier rules: Pot2-zero: ⊦ 2^ 0 = 1;%lemma-ax2: ⊦ 2^(n + 1) = 2 * 2^ n;%add: ⊦ m ^ (n + n0) = m ^ n * m ^ n0;%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%zero-02: ⊦ m * n = m * n0 ↔ m = 0 ∨ n = n0; 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