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Proof step 105 from proof hasSharing-dd-help (previous step)

# sks0 = 2 + # refs + # refs1, # sks = 2 + # refs0 + # refs2, fits(r, sks0, st), fits(r, sks, st), validrefs(str, (r ' + refs0) + r2 ' + refs2 + st[r -- sks] .refval ', st[r -- _type] .type, st), validrefs(str, (r ' + refs) + r1 ' + refs1 + st[r -- sks] .refval ', st[r -- _type] .type, st), ∀ n. n ≠ 0 ∧ ¬ # sks0 < n → st[r -- firstn(n, sks0)] .refval = ((r ' + refs) + r1 ' + refs1 + st[r -- sks] .refval ')[n], ∀ n. n ≠ 0 ∧ ¬ # sks < n → st[r -- firstn(n, sks)] .refval = ((r ' + refs0) + r2 ' + refs2 + st[r -- sks] .refval ')[n] ⊦ sks = [], sks0 = [], st[r -- sks] .refval = jvmref, r1 = st[r -- sks] .refval, r2 = r1, sks = sks0, st[r -- sks] = st[r -- sks0], path∃(str, r1, r2, st), path∃(str, r2, r1, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 104:
st[r -- sks] = refval(r0) ∧ is_referencevalue(st[r -- sks]), # sks0 = 2 + # refs + # refs1, # sks = 2 + # refs0 + # refs2, fits(r, sks0, st), fits(r, sks, st), validrefs(str, (r ' + refs0) + r2 ' + refs2 + r0 ', st[r -- _type] .type, st), validrefs(str, (r ' + refs) + r1 ' + refs1 + r0 ', st[r -- _type] .type, st), ∀ n. n ≠ 0 ∧ ¬ # sks0 < n → st[r -- firstn(n, sks0)] .refval = ((r ' + refs) + r1 ' + refs1 + r0 ')[n], ∀ n. n ≠ 0 ∧ ¬ # sks < n → st[r -- firstn(n, sks)] .refval = ((r ' + refs0) + r2 ' + refs2 + r0 ')[n] ⊦ sks = [], sks0 = [], r0 = jvmref, r1 = r0, r2 = r1, sks = sks0, st[r -- sks] = st[r -- sks0], path∃(str, r1, r2, st), path∃(str, r2, r1, st)

Rule argument

[speclemmaarg: is_referencevalue(val) ,r = val .refval ↔ val = refval(r) ,[val, r] → [st[r -- sks], r0] ,T ,javavalue , ,elim-refval ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ is_referencevalue(val) → val .refval = jvmref], [CSIMPELIM : is_referencevalue(val) ⊦ r = val .refval ↔ val = refval(r)])], [EXTRATERMS : (st[r -- sks], r0)]]