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Proof step 9 from proof potex (previous step)

2^ m < 2 * 2^ m, n0 < 2 * 2^ m, 2^ m ≤ n0 ⊦ n0 = 0, ∃ m0. 2^ m0 ≤ n0 + 1 ∧ n0 + 1 < 2 * 2^ m0

User applied rule exists right

And obtained the following new goals

Proof step 10:
n0 < 2 * 2^ m, 2^ m < 2 * 2^ m, 2^ m ≤ n0 ⊦ n0 = 0, ∃ m0. 2^ m0 ≤ n0 + 1 ∧ n0 + 1 < 2 * 2^ m0, 2^ m ≤ n0 + 1 ∧ n0 + 1 < 2 * 2^ m

Rule argument

pos: right 2 terms: m abort: F discard: F computed: F

Additional information

Used terms: m