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Proof step 11 from proof split-unique-2 (previous step)

n = m +1, 2^ n * 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)) ⊦ n = #fz(m0) + 1 ∧ n0 = lead(m0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
(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)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = m + n0 ↔ n = n0], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ 2^ (n + 1) = 2 * 2^ n], [CSIMPSEQ : ⊦ m +1 = m + 1])]]