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Proof step 7 from proof is_array-union-disjoint (previous step)

¬ ¬ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
st1 .refs ∩ st .refs = ∅, r ∈ st, is_array(r, st1), type∃(str, st1[r -- _type] .type), r ∈ st1, r -- _type ∈ st, r ∈ st .refs, ¬ 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 = abstract_type, st1[r -- _type] .type .jtelemtype = void_type, st = ∅

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ st .refs = ∅ ↔ st = ∅], [CSIMPFORWARD : ⊦ r ∈ rset → rset ≠ ∅], [CSIMPSEQ : ⊦ rset ∩ rset0 = rset0 ∩ rset], [CSIMPSEQ : ⊦ (rset ∩ rset0) ∩ rset1 = rset ∩ rset0 ∩ rset1])]]