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Proof step 5 from proof countdims-zero-04 (previous step)

0 < i0, 0 < i, countdims(ints0, i) ≤ 0, countdims(ints0, i) < i, ∀ i1. ¬ i1 ≤ 0 → 0 < countdims(ints0, i1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 < i, 0 < i0, countdims(ints0, i) ≤ 0, countdims(ints0, i) < i0, countdims(ints0, i) < i, ∀ i1. ¬ i1 ≤ 0 → 0 < countdims(ints0, i1) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPFORWARD : ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0], [CSIMPFORWARD : ⊦ i ≤ j ∧ j < j0 → i < j0])]]