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Proof step 6 from proof potmonleq (previous step)

∃ m0. 2^ n + m0 = 2^ m ∧ m0 ≠ 0, 2^ m ≤ 2^ n, n < m ⊦ n = m, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
2^ n + m0 = 2^ m, 2^ m ≤ 2^ n, n < m ⊦ n = m, m = 0, m0 = 0

Rule argument

Additional information