Proof step 6 from proof is_array-subset (conclusion)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 1:

  ⊦ is_array(str, r, st) ↔ ¬ ¬ ( type∃(str, st[r -- _type] .type) ∧ st[r -- _type] .type .jtelemtype ≠ void_type ∧ st[r -- _type] .type .jtelemtype ≠ abstract_type ∧ is_array(r, st))

• Proof step 2:

  ⊦ st ⊆ st1 ∧ is_array(r, st) → is_array(r, st1)

• Proof step 5:

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

Rule argument

Additional information

[[CONCRETESIMPRULES : ()]]