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Proof step 6 from proof countdims-zero-03 (previous step)

∀ i. ¬ (countdims(ints0, i) = 0 ∧ 0 < i), countdims(i0 + ints0, i) = 0, 0 < i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
countdims(i0 ' + ints0, i) = 0, 0 < i, ∀ i1. i1 ≤ 0 ∨ countdims(ints0, i1) ≠ 0 ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < i0 ∧ i = i1 → i1 < i0], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [CSIMPSEQ : ⊦ i + ints = i ' + ints])]]