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Proof step 3 from proof countrefs-i2n-02 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < i ⊦ i→n(1 + countrefs(ints) * i) = 1 + i→n(countrefs(ints)) * i→n(i)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j])]]