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Proof step 7 from proof finz-bounded-05 (previous step)

n0 = m +1, 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) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m ≤ #fz(m0), m0 + 1 < 2^ m * n + 2^ m, 2^ m * n ≤ m0, ∀ 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 : ⊦ m < n → n ≠ 0], [CSIMPFORWARD : ⊦ m < n → n ≠ 0], [CSIMPSEQ : ⊦ m + n ≤ m + n0 ↔ n ≤ n0], [CSIMPSEQ : ⊦ m * n + n < m1 * n + m2 * n ↔ ¬ ¬ (m + 1 < m1 + m2 ∧ n ≠ 0)], [CSIMPSEQ : ⊦ m * n = n * m], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n], [CSIMPSEQ : ⊦ 2 * n ≤ 2 * m + 1 ↔ n ≤ m], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ 2^ (n + 1) = 2 * 2^ n], [CSIMPSEQ : ⊦ m +1 = m + 1])]]