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Proof step 52 from proof addarrays-at-other-01 (previous step)

posints(ints), r ∈ st ⊦ (addarrays(newref_list(i→n(countrefs(ints)), st), ty, ints, i, st)[r -- sk]) = st[r -- sk], ints ≠ [] ∧ ¬ i < 0

User applied rule case distinction

And obtained the following new goals

Proof step 5:
posints(ints), r ∈ st ⊦ ints ≠ [], (addarrays(newref_list(i→n(countrefs(ints)), st), ty, ints, i, st)[r -- sk]) = st[r -- sk]
Proof step 51:
posints(ints), r ∈ st ⊦ ¬ i < 0, (addarrays(newref_list(i→n(countrefs(ints)), st), ty, ints, i, st)[r -- sk]) = st[r -- sk]

Rule argument

[CASEDARG : pos: right 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]