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Proof step 6 from proof inj (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: m0

Additional information

Used simplifier rules: 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 Used terms: m0