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Proof step 5 from proof countdims-notneg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i < 0 ∧ 0 ≤ j → i < j], [CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPFORWARD : ⊦ j < i ∧ i ≤ 0 → j < 0], [CSIMPFORWARD : ⊦ i < j ∧ j ≤ j0 → i < j0], [CSIMPFORWARD : ⊦ i < i0 ∧ i0 < i1 → i < i1])]]