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Proof step 4 from proof finz-bounded-02 (previous step)

m + 1 = 2^ n, m + 1 = 2^ #fz(m) * lead(m), m < 2^ n ⊦ n = #fz(m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m + 1 = 2^ n, m + 1 = 2^ #fz(m) * lead(m) ⊦ n = #fz(m)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ n < n], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1)], [CSIMPSEQ : ⊦ m + 1 ≠ 0], [CSIMPFORWARD : ⊦ m < n → n ≠ 0], [CSIMPFORWARD : ⊦ m < n ∧ n = n0 → m < n0], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPFORWARD : ⊦ m = n + 1 ∧ n < n0 → m ≤ n0], [CSIMPSEQ : ⊦ m + n = n + m])]]