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Proof step 4 from proof finz-bounded-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 2^ n ≠ 0], [CSIMPFORWARD : ⊦ m < n → n ≠ 0], [CSIMPSEQ : ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n], [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])]]