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Proof step 4 from proof split-unique-2 (previous step)

2^ n * n0 = 2 * m0 + 1, odd(n0), ∀ m1, n, n0. m1 < m0 + m0 → (2^ n * n0 = 2^ #fz(m1) * lead(m1) ∧ odd(n0) → n = #fz(m1) ∧ n0 = lead(m1)) ⊦ n = 0 ∧ n0 = 2 * m0 + 1

User applied rule constructor cut

And closed the goal!

Rule argument

[var-arg: n]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : nat generated by 0, +1;], [CSIMPSEQ : ⊦ m0 * 1 = m0], [CSIMPSEQ : ⊦ 2^ 0 = 1], [CSIMPSEQ : ⊦ m +1 = m + 1], [CSIMPSEQ : ⊦ 2^ (n + 1) = 2 * 2^ n], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ n0 < n → ( m * n + n0 = m0 * n ↔ m = m0 ∧ n0 = 0)], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n], [CSIMPSEQ : ⊦ m * n = n * m])]]