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Proof step 17 from proof finz-bounded (previous step)

2^ n + n0 = 2^ #fz(m), m + 1 = (2^ n + n0) * lead(m), n < #fz(m), ∃ m0. m + m0 = 2^ n ∧ m0 ≠ 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
(m + m0) + n0 = 2^ #fz(m), m + 1 = ((m + m0) + n0) * lead(m), m + m0 = 2^ n, n < #fz(m) ⊦ m0 = 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1])]]