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Proof step 5 from proof inj (previous step)

∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
2 * 2^ n0 = 2 * 2^ m0, ∀ n0. 2 * 2^ n0 = 2^ m0 ∧ (∀ m. 2^ n0 = 2^ m → n0 = m) → n0 + 1 = m0, ∀ m. 2^ n0 = 2^ m → n0 = m ⊦ n0 = m0

Rule argument

Additional information

Used simplifier rules: lemma-ax2: ⊦ 2^(n + 1) = 2 * 2^ n;%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2