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Proof step 5 from proof countrefs-i2n (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
(i2 + 1) * countrefs(ints) < 0, 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 : ([CSIMPSEQ : ⊦ i < 0 ∧ 0 ≤ j → i < j], [CSIMPFORWARD : ⊦ j < i ∧ i ≤ 0 → j < 0], [CSIMPFORWARD : ⊦ i < j ∧ j ≤ j0 → i < j0], [CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i])]]