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Proof step 8 from proof is_array-union-disjoint (previous step)

st1 .refs ∩ st .refs = ∅, r -- _type ∈ st, r ∈ st1, type∃(str, st1[r -- _type] .type), is_array(r, st1), r ∈ st, ¬ type∃(str, st[r -- _type] .type) ∨ st[r -- _type] .type .jtelemtype = void_type ∨ st[r -- _type] .type .jtelemtype = abstract_type ⊦ st1[r -- _type] .type .jtelemtype = void_type, st1[r -- _type] .type .jtelemtype = abstract_type

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ¬ r ∈ st .refs → ¬ r -- sk ∈ st
Proof step 7:
¬ ¬ r ∈ st .refs, st1 .refs ∩ st .refs = ∅, r -- _type ∈ st, r ∈ st1, type∃(str, st1[r -- _type] .type), is_array(r, st1), r ∈ st, ¬ type∃(str, st[r -- _type] .type) ∨ st[r -- _type] .type .jtelemtype = void_type ∨ st[r -- _type] .type .jtelemtype = abstract_type ⊦ st1[r -- _type] .type .jtelemtype = void_type, st1[r -- _type] .type .jtelemtype = abstract_type

Rule argument

[speclemmaarg: ,¬ r ∈ st .refs → ¬ r -- sk ∈ st ,[st, sk, r] → [st, _type, r] ,F ,storefunsx , ,refs-notin ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, _type, r)]]