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Proof step 7 from proof countrefs-i2n-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ ¬ countrefs(ints) < 0
Proof step 6:
0 ≤ countrefs(ints), 0 < i ⊦ i→n(countrefs(ints) * i) = i→n(countrefs(ints)) * i→n(i), i = 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPSEQ : ⊦ i ≠ j → (i ≤ j ↔ i < j)], [CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i], [CSIMPSEQ : ⊦ m * n = n * m], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n], [CSIMPSEQ : ⊦ m * 0 = 0], [CSIMPSEQ : ⊦ i * 0 = 0], [CSIMPSEQ : ⊦ i * j = j * i], [CSIMPSEQ : ⊦ i * j * j0 = (i * j) * j0], [CSIMPSEQ : ⊦ 0 ≤ i → (0 ≤ i * j ↔ ¬ ¬ (i = 0 ∨ 0 ≤ j))], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = m + n0 ↔ n = n0])]]