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Proof step 6 from proof getarray-subset (previous step)

j0 < j, st ⊆ st0, 0 ≤ i, 0 ≤ j0, 0 < j, ∀ j0. ¬ j0 < 0 ∧ ¬ i + j ≤ j0 → r -- j0 ' ∈ st ⊦ r -- (i + j0) ' ∈ st ∧ st ⊆ st0, st[r -- (i + j0) '] = st0[r -- (i + j0) ']

User applied rule case distinction

And obtained the following new goals

Proof step 5:
j0 < j, st ⊆ st0, 0 ≤ i, 0 ≤ j0, 0 < j, ∀ j0. ¬ j0 < 0 ∧ ¬ i + j ≤ j0 → r -- j0 ' ∈ st ⊦ r -- (i + j0) ' ∈ st, st[r -- (i + j0) '] = st0[r -- (i + j0) ']

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]