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Proof step 25 from proof validrefs-delete-03 (previous step)

st[r0 -- i '] = refval(r2), 0 ≤ i, 0 < st[r0 -- _length] .intval, is_array(str, r0, st), st[r0 -- _type] .type ≤ ty, i < st[r0 -- _length] .intval, is_referencevalue(st[r0 -- i ']), validrefs(str, r2 ' + refs, st[r0 -- _type] .type .type, st), ∀ str0, ty0, st0, r1. validrefs(str0, r2 ' + refs, ty0, st0) → r1 = r2 ∨ r1 ∈ refs ∨ validrefs(str0, r2 ' + refs, ty0, st0 -- r1) ⊦ r = r0, r0 = jvmref, r = r2, r ∈ refs, ∃ i0. ¬ i0 < 0 ∧ ¬ st[r0 -- _length] .intval ≤ i0 ∧ is_referencevalue(st[r0 -- i0 ']) ∧ r2 = st[r0 -- i0 '] .refval ∧ validrefs(str, r2 ' + refs, st[r0 -- _type] .type .type, st -- r)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
st[r0 -- i '] = refval(r2), validrefs(str, r2 ' + refs, st[r0 -- _type] .type .type, st), i < st[r0 -- _length] .intval, st[r0 -- _type] .type ≤ ty, is_array(str, r0, st), 0 < st[r0 -- _length] .intval, 0 ≤ i, ∀ str0, ty0, st0, r1. validrefs(str0, r2 ' + refs, ty0, st0) → r1 = r2 ∨ r1 ∈ refs ∨ validrefs(str0, r2 ' + refs, ty0, st0 -- r1) ⊦ r = r2, r0 = jvmref, r = r0, r ∈ refs, ∃ i0. ¬ i0 < 0 ∧ ¬ st[r0 -- _length] .intval ≤ i0 ∧ is_referencevalue(st[r0 -- i0 ']) ∧ r2 = st[r0 -- i0 '] .refval ∧ validrefs(str, st[r0 -- i0 '] .refval ' + refs, st[r0 -- _type] .type .type, st -- r)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ is_referencevalue(refval(r))], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1])]]