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Proof step 1 from proof inj-02 (conclusion)

⊦ 2^ n = 2 * 2^ m ↔ n = m + 1

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ 2^ n = 2^ m ↔ n = m
Proof step 3:
2^ n = 2^(m + 1) ↔ n = m + 1 ⊦ 2^ n = 2 * 2^ m ↔ n = m + 1

Rule argument

spec: <toplevel> name: inj seq: ⊦ 2^ n = 2^ m ↔ n = m subst: [n, m] → [n, m + 1]

Additional information

Used terms: n, m + 1