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Proof step 15 from proof multi-subset (previous step)

# 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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ st1 ⊆ st2 ↔ (∀ rk. rk ∈ st1 → rk ∈ st2 ∧ st1[rk] = st2[rk])
Proof step 14:
# refs = i→n(countrefs(i ' + ints0)), i ≤ i0, 0 ≤ i, is_newref_list(refs, st), posints(ints0), 0 ≤ i0 ⊦ ∀ rk. rk ∈ st → rk ∈ addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st) ∧ st[rk] = (addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st)[rk]), 0 ∈ (i0 ' + ints0) .butlast

Rule argument

[speclemmaarg: ,st1 ⊆ st2 ↔ (∀ rk. rk ∈ st1 → rk ∈ st2 ∧ st1[rk] = st2[rk]) ,[st2, st1] → [addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st), st] ,F ,store ,javastore ,Subset ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st), st)]]