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Proof step 9 from proof split-ok (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ #fz(2 * n + 1) = #fz(n) + 1
Proof step 7:
⊦ lead(2 * n + 1) = lead(n)
Proof step 8:
∀ n0. n0 < m + m + 1 → (true → n0 + 1 = lead(n0) * 2 ^ #fz(n0)) ⊦ m + 1 = 2 ^ #fz(m) * lead(m)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m * n + n = m1 * n ↔ ¬ ¬ (m + 1 = m1 ∨ n = 0)], [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])]]