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Proof step 26 from proof validrefs-addarray (previous step)

0 ≤ i, 0 < st[r0 -- _length] .intval, newref(r, st), st[r0 -- _type] .type ≤ ty, is_array(str, r0, st), i < st[r0 -- _length] .intval, is_referencevalue(st[r0 -- i ']), validrefs(str, st[r0 -- i '] .refval ' + refs, st[r0 -- _type] .type .type, addarray(r, ty0, vals, st)), ∀ str0, ty1, r1, ty2, vals0, st0. validrefs(str0, st[r0 -- i '] .refval ' + refs, ty1, addarray(r1, ty2, vals0, st0)) ∧ newref(r1, st0) → r1 = jvmref ∨ r1 = st[r0 -- i '] .refval ∨ r1 ∈ refs ∨ validrefs(str0, st[r0 -- i '] .refval ' + refs, ty1, st0) ⊦ r = jvmref, r = r0, r0 = jvmref, 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 ∧ validrefs(str, st[r0 -- i '] .refval ' + refs, st[r0 -- _type] .type .type, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 25:
st[r0 -- i '] = refval(r2), 0 ≤ i, 0 < st[r0 -- _length] .intval, newref(r, st), st[r0 -- _type] .type ≤ ty, is_array(str, r0, st), i < st[r0 -- _length] .intval, is_referencevalue(st[r0 -- i ']), validrefs(str, r2 ' + refs, st[r0 -- _type] .type .type, addarray(r, ty0, vals, st)), ∀ str0, ty1, r1, ty2, vals0, st0. validrefs(str0, r2 ' + refs, ty1, addarray(r1, ty2, vals0, st0)) ∧ newref(r1, st0) → r1 = jvmref ∨ r1 = r2 ∨ r1 ∈ refs ∨ validrefs(str0, r2 ' + refs, ty1, st0) ⊦ r = jvmref, 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)

Rule argument

[speclemmaarg: ,is_referencevalue(val) → (r = val .refval ↔ val = refval(r)) ,[val, r] → [st[r0 -- i '], r2] ,T ,javavalue , ,elim-13 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_referencevalue(val) → ( r = val .refval ↔ val = refval(r))])], [EXTRATERMS : (st[r0 -- i '], r2)]]