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Proof step 14 from proof split-unique (previous step)

∃ m. m + m = n, lead(n) * 2 ^ #fz(n) = lead(m) * 2 ^ #fz(m), ∀ 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 2:
⊦ #fz(2 * n) = 0
Proof step 3:
⊦ lead(2 * n) = 2 * n + 1
Proof step 13:
2 * m0 + 1 = lead(m) * 2 ^ #fz(m), ∀ n0, m. n0 < m0 + m0 → (lead(n0) * 2 ^ #fz(n0) = lead(m) * 2 ^ #fz(m) → m = n0) ⊦ m = 2 * m0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m0 * 1 = m0], [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])]]