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Proof step 4 from proof countrefs2-03 (previous step)

¬ 0 < countrefs2(ints) ⊦ i→n(countrefs2(ints) + 1) = i→n(countrefs2(ints)) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
countrefs2(ints) ≤ 0 ⊦ i→n(countrefs2(ints) + 1) = i→n(countrefs2(ints)) + 1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i])]]