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Proof step 15 from proof finz-bounded-05 (conclusion)

⊦ n * 2^ n0 ≤ m ∧ m + 1 < (n + 1) * 2^ n0 → #fz(m) < n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
n0 ≤ #fz(m), m + 1 < 2^ n0 * n + 2^ n0, n * 2^ n0 ≤ m ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m < n → n ≠ 0], [CSIMPFORWARD : ⊦ m < n → n ≠ 0], [CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m], [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 * (n + 1) = m * n + m])]]