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Proof step 9 from proof finz-bounded (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
1 + m = 2^ #fz(m), m + 1 = 2^ n, lead(m) = 1, n < #fz(m) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : n < n0 ⊦ ((* n)) ≠ m0 + ((* n0))], [CSIMPSEQ : ⊦ m + n = n + m])]]