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Proof step 12 from proof multi-refin-other (previous step)

# refs = i→n(countrefs(i ' + ints0)), i ≤ i0, 0 ≤ i, posints(ints0), 0 ≤ i0 ⊦ (∃ sk. r -- sk ∈ addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st)) ↔ (∃ sk. r -- sk ∈ st), r ∈ refs, 0 ∈ (i0 ' + ints0) .butlast

User applied rule case distinction

And obtained the following new goals

Proof step 7:
∃ sk. r -- sk ∈ addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st), # refs = i→n(countrefs(i ' + ints0)), i ≤ i0, 0 ≤ i, posints(ints0), 0 ≤ i0 ⊦ ∃ sk. r -- sk ∈ st, r ∈ refs, 0 ∈ (i0 ' + ints0) .butlast
Proof step 11:
∃ sk. r -- sk ∈ st, # refs = i→n(countrefs(i ' + ints0)), i ≤ i0, 0 ≤ i, posints(ints0), 0 ≤ i0 ⊦ ∃ sk. r -- sk ∈ addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st), r ∈ refs, 0 ∈ (i0 ' + ints0) .butlast

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]