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Proof step 6 from proof isarray-02 (previous step)

i ≤ j ⊦ (∀ j0. ¬ j0 < 0 ∧ ¬ i ≤ j0 → r -- j0 ' ∈ st[r0 -- j ', val]) ↔ (∀ j. 0 ≤ j ∧ j < i → r -- j ' ∈ st)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ j0. ¬ j0 < 0 ∧ ¬ i ≤ j0 → r -- j0 ' ∈ st[r0 -- j ', val], i ≤ j ⊦ ∀ j. 0 ≤ j ∧ j < i → r -- j ' ∈ st
Proof step 5:
∀ j. 0 ≤ j ∧ j < i → r -- j ' ∈ st, i ≤ j ⊦ ∀ j0. ¬ j0 < 0 ∧ ¬ i ≤ j0 → r -- j0 ' ∈ st[r0 -- j ', val]

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]