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Proof step 11 from proof finz-bounded (previous step)

(m + m0) + n0 = 2^ #fz(m), m + 1 = ((m + m0) + n0) * 2 * m1 + ((m + m0) + n0) * 1, m + m0 = 2^ n, lead(m) = 2 * m1 + 1, n < #fz(m) ⊦ m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
lead(m) = 2 * m1 + 1, m + 1 = 2^ n, (2 + 2 * m) * m1 = 0, 1 + m = 2^ #fz(m), n < #fz(m) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 + m = m], [CSIMPSEQ : ⊦ m + n = 0 ↔ m = 0 ∧ n = 0], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m * (n + 1) = m * n + m], [CSIMPSEQ : ⊦ n ≠ 0 → (m + n = 1 ↔ n = 1 ∧ m = 0)], [CSIMPSEQ : ⊦ m + n = m + n0 ↔ n = n0], [CSIMPSEQ : ⊦ m * n = n * m], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n], [CSIMPSEQ : ⊦ m0 * 1 = m0])]]