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Proof step 4 from proof getarray-subset-01 (conclusion)

⊦ st ⊆ st0 ∧ is_array(str, r, st) → getarray(r, st) = getarray(r, st0)

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 3:
st ⊆ st0, type∃(str, st[r -- _type] .type), is_array(r, st) ⊦ getarray(r, st) = getarray(r, st0), st[r -- _type] .type .jtelemtype = abstract_type, st[r -- _type] .type .jtelemtype = void_type

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]