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Proof step 16 from proof multi-subset (previous step)

ints = i0 + ints0, # refs = i→n(countrefs(i ' + ints0)), is_newref_list(refs, st), posints(ints), i ≤ i0, 0 ≤ i, 0 ≤ i0 ⊦ ints = [], st ⊆ addmultiarrays(refs, i, acc, ty, ints, st), 0 ∈ ints .butlast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# refs = i→n(countrefs(i ' + ints0)), i ≤ i0, 0 ≤ i, is_newref_list(refs, st), posints(ints0), 0 ≤ i0 ⊦ st ⊆ addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st), 0 ∈ (i0 ' + ints0) .butlast

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPSEQ : ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j], [CSIMPSEQ : ⊦ posints(i ' + ints) ↔ 0 ≤ i ∧ posints(ints)], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ i ' + ints ≠ []], [CSIMPSEQ : ⊦ i + ints = i ' + ints])]]