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Proof step 4 from proof potmonleq (previous step)

2^ n < 2^ m, 2^ m ≤ 2^ n, n < m ⊦ n = m, m = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0)
Proof step 6:
∃ m0. 2^ n + m0 = 2^ m ∧ m0 ≠ 0, 2^ m ≤ 2^ n, n < m ⊦ n = m, m = 0

Rule argument

spec: nat-basic2 inst: name: addex seq: ⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0) subst: [n, m] → [2^ m, 2^ n] flags: rewrite

Additional information

Used terms: 2^ m, 2^ n