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Proof step 15 from proof odd-add (previous step)

⊦ 2 * m / 2 = m, ∃ m0. 2 * m0 = 2 * m1 + m + 1

User applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
⊦ odd(n) → (m = n / 2 ↔ n = m + m + 1)
Proof step 17:
⊦ 2 * m / 2 = m, odd(m), ∃ m0. 2 * m0 = 2 * m1 + m + 1
Proof step 19:
m = m2 + m2 + 1, odd(m) ⊦ 2 * m2 = m, ∃ m0. 2 * m0 = 2 * m1 + m + 1

Rule argument

spec: <toplevel> name: div2-elim seq: ⊦ odd(n) → (m = n / 2 ↔ n = m + m + 1) subst: [n, m] → [m, m2]

Additional information

Used terms: m, m2