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Proof step 4 from proof finz-bigger (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ #fz(2 * n) = 0
Proof step 3:
2 * m0 + 1 = n * 2^ n0, ∀ m1, n, n0. m1 < m0 + m0 → (m1 + 1 = n * 2^ n0 ∧ #fz(m1) < n0 → false) ⊦ n0 = 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 < m ↔ m ≠ 0], [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])]]