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Proof step 51 from proof junkrefs-subset (previous step)

acc = r0 + refs, r0 = st1[r -- i '] .refval, st1[r -- _type] .type ≤ ty, is_array(str, r, st), i < st1[r -- _length] .intval, is_referencevalue(st1[r -- i ']), junkrefs(str, acc, st1[r -- _type] .type .type, st1), st ⊆ st1, st[r -- _type] .type ≤ ty, 0 ≤ i, 0 < st1[r -- _length] .intval, ∀ 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 ']) ∧ r0 = 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 50:
st ⊆ st1, st[r -- _type] .type ≤ ty, 0 ≤ i, 0 < st1[r -- _length] .intval, st1[r -- _type] .type ≤ ty, is_array(str, r, st), i < st1[r -- _length] .intval, is_referencevalue(st1[r -- i ']), junkrefs(str, st1[r -- i '] .refval ' + refs, st1[r -- _type] .type .type, st1), ∀ str0, ty0, st0, st2. junkrefs(str0, st1[r -- i '] .refval ' + refs, ty0, st0) ∧ st2 ⊆ st0 → junkrefs(str0, st1[r -- i '] .refval ' + refs, ty0, st2) ⊦ r = jvmref, ∃ i0. ¬ i0 < 0 ∧ ¬ st[r -- _length] .intval ≤ i0 ∧ is_referencevalue(st[r -- i0 ']) ∧ st1[r -- i '] .refval = st[r -- i0 '] .refval ∧ junkrefs(str, st1[r -- i '] .refval ' + refs, st[r -- _type] .type .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' + refs ≠ []], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]