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Proof step 9 from proof ls-01 (previous step)

m1 * (n + n2) + n1 < m1 * n, n1 < n + n2, m < m1 * n, m1 * (n + n2) + n1 ≤ m ⊦ m1 = 0, n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ m0 * (m + n) = m0 * m + m0 * n

Rule argument

spec: nat-mult name: dist seq: ⊦ m0 * (m + n) = m0 * m + m0 * n subst: [n, m, m0] → [n2, n, m1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n2, n, m1