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Proof step 10 from proof split-unique-2 (previous step)

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

User applied rule apply induction

And closed the goal!

Rule argument

[substlist-arg: [m1, n, n0] → [m0, m, n0]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m * n = m * n0 ↔ m = 0 ∨ n = n0], [CSIMPSEQ : ⊦ m + m = 2 * m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ ¬ 2 * n < n], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n], [CSIMPSEQ : ⊦ m * n = n * m])], [EXTRATERMS : (m0, m, n0)]]