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Proof step 7 from proof multi-isarrayref-notin (previous step)

∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# refs = i→n(countrefs(i ' + ints0)), j0 < j, i ≤ i0, 0 ≤ i, posints(ints0), 0 ≤ i0, 0 ≤ j0, 0 < j, ∀ j0. ¬ j0 < 0 ∧ ¬ j ≤ j0 → r -- j0 ' ∈ addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st) ⊦ r -- j0 ' ∈ st, r ∈ refs, 0 ∈ (i0 ' + ints0) .butlast

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j])]]