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Proof step 30 from proof even-add (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 31:
⊦ odd(n) → (m = n / 2 ↔ n = m + m + 1)
Proof step 32:
⊦ odd(n), (2 * m / 2 + 2 * n / 2) + 2 = m + n, ∃ m0. 2 * m0 = n, ∃ m0. 2 * m0 = m
Proof step 34:
n = m1 + m1 + 1 ∧ odd(n) ⊦ (2 * m / 2 + 2 * m1) + 2 = m + n, ∃ m0. 2 * m0 = n, ∃ m0. 2 * m0 = m

Rule argument

spec: nat-even inst: name: div2-elim seq: ⊦ odd(n) → (m = n / 2 ↔ n = m + m + 1) subst: [n, m] → [n, m1] flags: current, []

Additional information

Used terms: n, m1