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Proof step 11 from proof lead-is-odd (previous step)

true, ∀ n0. n0 < n → (true → odd(lead(n0))) ⊦ odd(lead(n))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
even(lead(n)), ∀ n0. n0 < n → (true → odd(lead(n0))) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ odd(n) ↔ even(n)])]]