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Proof step 27 from proof junkrefs-delete (previous step)

acc = r1 + refs, r1 = st[r0 -- i '] .refval, st[r0 -- _type] .type ≤ ty, is_array(str, r0, st), i < st[r0 -- _length] .intval, is_referencevalue(st[r0 -- i ']), junkrefs(str, acc, st[r0 -- _type] .type .type, st -- r), 0 ≤ i, 0 < st[r0 -- _length] .intval, ∀ str, ty, st, r. junkrefs(str, acc, ty, st -- r) → r ∈ acc ∨ junkrefs(str, acc, ty, st) ⊦ r0 = jvmref, r = r0, acc = [], r ∈ acc, ∃ i. ¬ i < 0 ∧ ¬ st[r0 -- _length] .intval ≤ i ∧ is_referencevalue(st[r0 -- i ']) ∧ r1 = st[r0 -- i '] .refval ∧ junkrefs(str, acc, st[r0 -- _type] .type .type, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
0 ≤ i, 0 < st[r0 -- _length] .intval, st[r0 -- _type] .type ≤ ty, is_array(str, r0, st), i < st[r0 -- _length] .intval, is_referencevalue(st[r0 -- i ']), junkrefs(str, st[r0 -- i '] .refval ' + refs, st[r0 -- _type] .type .type, st -- r), ∀ str0, ty0, st0, r1. junkrefs(str0, st[r0 -- i '] .refval ' + refs, ty0, st0 -- r1) → r1 = st[r0 -- i '] .refval ∨ r1 ∈ refs ∨ junkrefs(str0, st[r0 -- i '] .refval ' + refs, ty0, st0) ⊦ r0 = jvmref, r = r0, r = st[r0 -- i '] .refval, r ∈ refs, ∃ i0. ¬ i0 < 0 ∧ ¬ st[r0 -- _length] .intval ≤ i0 ∧ is_referencevalue(st[r0 -- i0 ']) ∧ st[r0 -- i '] .refval = st[r0 -- i0 '] .refval ∧ junkrefs(str, st[r0 -- i '] .refval ' + refs, st[r0 -- _type] .type .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ∈ r0 ' ↔ r = r0], [CSIMPSEQ : ⊦ r ∈ refs + refs0 ↔ ¬ ¬ (r ∈ refs ∨ r ∈ refs0)], [CSIMPSEQ : ⊦ ¬ r ∈ r0 ' + refs ↔ r ≠ r0 ∧ ¬ r ∈ refs], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' + refs ≠ []], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]