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Proof step 27 from proof split-unique (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
⊦ #fz(2 * n + 1) = #fz(n) + 1
Proof step 25:
⊦ lead(2 * n + 1) = lead(n)
Proof step 26:
(2 ^ #fz(m0) * 2) * lead(m0) = (2 ^ #fz(m1) * 2) * lead(m1), ∀ n0, m. n0 < m0 + m0 + 1 → (lead(n0) * 2 ^ #fz(n0) = lead(m) * 2 ^ #fz(m) → m = n0) ⊦ m1 = m0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m * n = m * n0 ↔ m = 0 ∨ n = n0], [CSIMPSEQ : ⊦ m + n = m + n0 ↔ n = n0], [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])]]