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Proof step 4 from proof inj (previous step)

2 * 2^ n0 = 2^ m, ∀ m. 2^ n0 = 2^ m → n0 = m ⊦ n0 + 1 = m

User applied rule structural induction

And obtained the following new goals

Proof step 5:
∀ n0. 2 * 2^ n0 = 2^ m0 ∧ (∀ m. 2^ n0 = 2^ m → n0 = m) → n0 + 1 = m0, 2 * 2^ n0 = 2^ m0 +1, ∀ m. 2^ n0 = 2^ m → n0 = m ⊦ n0 + 1 = m0 +1

Rule argument

indvar: m

Additional information

Used simplifier rules: Pot2-zero: ⊦ 2^ 0 = 1;%nat freely generated by 0, +1;;%one-01: ⊦ m0 * m = 1 ↔ m0 = 1 ∧ m = 1; 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