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Proof step 31 from proof split-unique (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
⊦ #fz(2 * n + 1) = #fz(n) + 1
Proof step 18:
⊦ lead(2 * n + 1) = lead(n)
Proof step 30:
(2 ^ #fz(m0) * 2) * lead(m0) = lead(m) * 2 ^ #fz(m), ∀ n0, m. n0 < m0 + m0 + 1 → (lead(n0) * 2 ^ #fz(n0) = lead(m) * 2 ^ #fz(m) → m = n0) ⊦ m = 2 * m0 + 1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m ^ (n + n0) = m ^ n * m ^ n0], [CSIMPSEQ : ⊦ m * n = n * m], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + m = 2 * m])]]