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Proof step 81 from proof pseq-subset (previous step)

st1[r1 -- _type] .type = mkclasstype(class), st[r -- _type] = st1[r1 -- _type], st2[r -- _type] = st1[r1 -- _type], r ∈ st2, r ∈ st, r -- _type ∈ st2, r -- _type ∈ st, is_referencevalue(st2[r -- fs ']), fs ' ∈ instfields(class), ∀ rk. rk ∈ st → rk ∈ st2 ∧ st[rk] = st2[rk], ∀ fs. fs ' ∈ instfields(class) ∧ is_referencevalue(st[r -- fs ']) → ps=(st[r -- fs '] .refval, st, st1[r1 -- fs '] .refval, st1), ∀ fs. fs ' ∈ instfields(class) ∧ ¬ is_referencevalue(st[r -- fs ']) → st[r -- fs '] = st1[r1 -- fs '], ∀ st1, r1, st2, st0, r0, str0. maxpath(str0, r0, st0) < maxpath(str, r, st) → ( st0 ⊆ st2 ∧ ps=(r0, st0, r1, st1) → junkref(str0, r0, st0) ∨ cyclic(str0, r0, st0) ∨ junkref(str0, r1, st1) ∨ cyclic(str0, r1, st1) ∨ ps=(r0, st2, r1, st1)) ⊦ r = jvmref, r1 = jvmref, junkref(str, r, st), cyclic(str, r, st), junkref(str, r1, st1), cyclic(str, r1, st1), ps=(st2[r -- fs '] .refval, st2, st1[r1 -- fs '] .refval, st1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

[speclemmaarg: ,is_referencevalue(val) → (r = val .refval ↔ val = refval(r)) ,[val, r] → [st2[r -- fs '], r2] ,T ,javavalue , ,elim-13 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_referencevalue(val) → ( r = val .refval ↔ val = refval(r))])], [EXTRATERMS : (st2[r -- fs '], r2)]]