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Proof step 36 from proof junkrefs-subset (previous step)

¬ (is_array(str, r, st1) ∧ st1[r -- _type] .type ≤ ty), is_array(str, r, st), junkrefs(str, r ' + acc, ty, st1), st ⊆ st1, st[r -- _type] .type ≤ ty, ∀ str, ty, st1, st. junkrefs(str, acc, ty, st1) ∧ st ⊆ st1 → junkrefs(str, acc, ty, st) ⊦ r = jvmref, acc ≠ [] ∧ (∃ i. ¬ i < 0 ∧ ¬ st[r -- _length] .intval ≤ i ∧ is_referencevalue(st[r -- i ']) ∧ acc .first = st[r -- i '] .refval ∧ junkrefs(str, acc, st[r -- _type] .type .type, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
is_array(str, r, st), junkrefs(str, r ' + acc, ty, st1), st ⊆ st1, st[r -- _type] .type ≤ ty, ∀ str, ty, st1, st. junkrefs(str, acc, ty, st1) ∧ st ⊆ st1 → junkrefs(str, acc, ty, st) ⊦ r = jvmref, st1[r -- _type] .type ≤ ty, acc ≠ [] ∧ (∃ i. ¬ i < 0 ∧ ¬ st[r -- _length] .intval ≤ i ∧ is_referencevalue(st[r -- i ']) ∧ acc .first = st[r -- i '] .refval ∧ junkrefs(str, acc, st[r -- _type] .type .type, st))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ st ⊆ st1 ∧ is_array(str, r, st) → is_array(str, r, st1)])]]