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Proof step 10 from proof potex (previous step)

2 * 2^ m = n0 + 1, 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 11:
2 * 2^ m = n0 + 1, 2^ m ≤ n0 ⊦ n0 = 0, 2^(m + 1) ≤ n0 + 1 ∧ n0 + 1 < 2 * 2^(m + 1), ∃ m0. 2^ m0 ≤ n0 + 1 ∧ n0 + 1 < 2 * 2^ m0

Rule argument

pos: right 2 terms: m + 1 abort: F discard: F computed: F

Additional information

Used terms: m + 1