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Proof step 9 from proof odd-add (previous step)

⊦ 2 * n / 2 = n, ∃ m0. 2 * m0 = m + n, ∃ m0. 2 * m0 = m

User applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
⊦ odd(n) → (m = n / 2 ↔ n = m + m + 1)
Proof step 11:
⊦ 2 * n / 2 = n, odd(n), ∃ m0. 2 * m0 = m, ∃ m0. 2 * m0 = m + n
Proof step 13:
n = m1 + m1 + 1, odd(n) ⊦ 2 * m1 = n, ∃ m0. 2 * m0 = m, ∃ m0. 2 * m0 = m + n

Rule argument

spec: <toplevel> name: div2-elim seq: ⊦ odd(n) → (m = n / 2 ↔ n = m + m + 1) subst: [n, m] → [n, m1]

Additional information

Used terms: n, m1