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Proof step 13 from proof split-ok (previous step)

true, ∀ n0. n0 < n → (true → n0 + 1 = lead(n0) * 2 ^ #fz(n0)) ⊦ n + 1 = lead(n) * 2 ^ #fz(n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ n0. n0 < n → (true → n0 + 1 = lead(n0) * 2 ^ #fz(n0)) ⊦ n + 1 = lead(n) * 2 ^ #fz(n)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]