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Proof step 19 from proof split-unique-2 (conclusion)

⊦ odd(n0) → (2^ n * n0 = 2^ #fz(m) * lead(m) ↔ n = #fz(m) ∧ n0 = lead(m))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
2^ n * n0 = 2^ #fz(m) * lead(m), odd(n0) ⊦ n = #fz(m) ∧ n0 = lead(m)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m * n = n * m])]]