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Proof step 11 from proof countrefs-i2n (conclusion)

⊦ 0 ≤ i2 → i→n(1 + (i2 + 1) * countrefs(ints)) = 1 + i→n(i2 + 1) * i→n(countrefs(ints))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
0 ≤ i2 ⊦ i→n(1 + (i2 + 1) * countrefs(ints)) = 1 + i→n(countrefs(ints)) * i→n(i2) + i→n(countrefs(ints))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [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 * (n + 1) = m * n + m], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1])]]