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Proof step 19 from proof hasSharing-dd-help (previous step)

st[r -- firstn(1 + # refs, sks0)] .refval = (refs + r1 ' + refs1 + r0 ')[# refs], # refs1 = # refs2, # refs = # refs0, # sks0 = 2 + # refs2 + # refs, validrefs(str, (r ' + refs) + r1 ' + refs1 + r0 ', st[r -- _type] .type, st), validrefs(str, (r ' + refs0) + (refs + r1 ' + refs1 + r0 ')[# refs] ' + refs2 + r0 ', st[r -- _type] .type, st), fits(r, sks0, st) ⊦ (refs + r1 ' + refs1 + r0 ')[# refs] = r1, r1 = r0, r0 = jvmref, path∃(str, (refs + r1 ' + refs1 + r0 ')[# refs], r1, st), path∃(str, r1, (refs + r1 ' + refs1 + r0 ')[# refs], st), hasSharing(r, st)

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

[speclemmaarg: is_referencevalue(val) ,r = val .refval ↔ val = refval(r) ,[val, r] → [st[r -- firstn(1 + # refs, sks0)], r2] ,T ,javavalue , ,elim-refval ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ is_referencevalue(val) → val .refval = jvmref], [CSIMPSEQ : ⊦ (refs + r ' + refs0)[# refs] = r], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPELIM : is_referencevalue(val) ⊦ r = val .refval ↔ val = refval(r)])], [EXTRATERMS : (st[r -- firstn(1 + # refs, sks0)], r2)]]