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Proof step 3 from proof isarray-03 (previous step)

⊦ (∀ j. ¬ j < 0 ∧ ¬ 0 < j → r -- j ' ∈ st) ↔ r -- 0 ' ∈ st

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 2:
∀ j. ¬ j < 0 ∧ ¬ 0 < j → r -- j ' ∈ st ⊦ r -- 0 ' ∈ st

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ i < i0 → (¬ i0 < i ↔ i0 = i)])]]