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Proof step 9 from proof is_array-union-disjoint (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
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

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st], [CSIMPSEQ : ⊦ rk ∈ st0 → (st ∪ st0)[rk] = st0[rk]])]]