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Proof step 7 from proof lead-is-odd (previous step)

even(lead(n)), ∃ m. n = m + m + 1, ∀ n0. n0 < n → (true → odd(lead(n0))) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ lead(2 * n + 1) = lead(n)
Proof step 6:
even(lead(m)), ∀ n0. n0 < m + m + 1 → (true → odd(lead(n0))) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m * n = n * m], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + m = 2 * m])]]