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Proof step 5 from proof is_array-union (previous step)

type∃(str, st[r -- _type] .type), is_array(r, st), ¬ type∃(str, (st0 ∪ st)[r -- _type] .type) ∨ (st0 ∪ st)[r -- _type] .type .jtelemtype = void_type ∨ (st0 ∪ st)[r -- _type] .type .jtelemtype = abstract_type ⊦ st[r -- _type] .type .jtelemtype = void_type, st[r -- _type] .type .jtelemtype = abstract_type

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ rk ∈ st0 → (st ∪ st0)[rk] = st0[rk]
Proof step 4:
⊦ ¬ r -- _type ∈ st → ¬ is_array(r, st)

Rule argument

[speclemmaarg: ,rk ∈ st0 → (st ∪ st0)[rk] = st0[rk] ,[rk, st0, st] → [r -- _type, st, st0] ,F ,storefunsx , ,union-at-right-01 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]], [CSIMPSEQ : ⊦ ∅[rk] = noval])], [EXTRATERMS : (r -- _type, st, st0)]]