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Proof step 1 from proof pot-01 (conclusion)

n ≠ 0 ⊦ 2 * 2^ n -1 = 2^ n

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ 2^(n + 1) = 2 * 2^ n
Proof step 3:
2^(n -1 + 1) = 2 * 2^ n -1, n ≠ 0 ⊦ 2 * 2^ n -1 = 2^ n

Rule argument

spec: nat-pot inst: name: lemma-ax2 seq: ⊦ 2^(n + 1) = 2 * 2^ n subst: [n] → [n -1] flags: current, []

Additional information

Used terms: n -1