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Proof step 13 from proof multi-isarrayref-notin (previous step)

# refs = i→n(countrefs(i ' + ints0)), i ≤ i0, 0 ≤ i, posints(ints0), 0 ≤ i0 ⊦ (∀ j0. ¬ j0 < 0 ∧ ¬ j ≤ j0 → r -- j0 ' ∈ addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st)) ↔ (∀ j0. 0 ≤ j0 ∧ j0 < j → r -- j0 ' ∈ st), r ∈ refs, 0 ∈ (i0 ' + ints0) .butlast

User applied rule case distinction

And obtained the following new goals

Proof step 7:
∀ j0. ¬ j0 < 0 ∧ ¬ j ≤ j0 → r -- j0 ' ∈ addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st), # refs = i→n(countrefs(i ' + ints0)), i ≤ i0, 0 ≤ i, posints(ints0), 0 ≤ i0 ⊦ ∀ j0. 0 ≤ j0 ∧ j0 < j → r -- j0 ' ∈ st, r ∈ refs, 0 ∈ (i0 ' + ints0) .butlast
Proof step 12:
∀ j0. 0 ≤ j0 ∧ j0 < j → r -- j0 ' ∈ st, # refs = i→n(countrefs(i ' + ints0)), i ≤ i0, 0 ≤ i, posints(ints0), 0 ≤ i0 ⊦ ∀ j0. ¬ j0 < 0 ∧ ¬ j ≤ j0 → r -- j0 ' ∈ addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st), r ∈ refs, 0 ∈ (i0 ' + ints0) .butlast

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]