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Proof step 9 from proof finz-bounded-05 (previous step)

n * 2^ n0 ≤ m, m + 1 < 2^ n0 * n + 2^ n0, n0 ≤ #fz(m), ∃ m0. m = m0 + m0 + 1, ∀ m1, n0, n. m1 < m → (n0 ≤ #fz(m1) ∧ m1 + 1 < 2^ n0 * n + 2^ n0 ∧ n * 2^ n0 ≤ m1 → false) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ #fz(2 * n + 1) = #fz(n) + 1
Proof step 8:
n * 2^ n0 ≤ 2 * m0 + 1, 2 * m0 + 2 < 2^ n0 * n + 2^ n0, n0 ≤ #fz(m0) + 1, ∀ m1, n0, n. m1 < m0 + m0 + 1 → (n0 ≤ #fz(m1) ∧ m1 + 1 < 2^ n0 * n + 2^ n0 ∧ n * 2^ n0 ≤ m1 → false) ⊦

Rule argument

[emptyarg]

Additional information

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