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Proof step 87 from proof cyclic-pseq (previous step)

st[r2 -- _type] = st1[r1 -- _type], st1[r1 -- _type] .type = mkclasstype(class), is_referencevalue(st[r2 -- fs ']), fs ' ∈ instfields(class), is_obj(str, r1, st1), is_obj(str, r2, st), validrefs(str, (st[r2 -- fs '] .refval ' + refs) + r0 ', fs .type, st), ps=(st[r2 -- fs '] .refval, st, st1[r1 -- fs '] .refval, st1), ∀ fs. fs ' ∈ instfields(class) ∧ ¬ is_referencevalue(st[r2 -- fs ']) → st[r2 -- fs '] = st1[r1 -- fs '], ∀ st, r, st0, r0, str0. maxpath(str0, r0, st0) < maxpath(str, r1, st1) → (ps=(r, st, r0, st0) ∧ cyclic(str0, r, st) → junkref(str0, r, st) ∨ junkref(str0, r0, st0) ∨ cyclic(str0, r0, st0)) ⊦ r1 = jvmref, r2 = jvmref, r2 = st[r2 -- fs '] .refval, junkref(str, r2, st), junkref(str, r1, st1), cyclic(str, r1, st1), r2 ∈ refs, dups(refs), st[r2 -- fs '] .refval ∈ refs, ∃ refs. dups(refs) ∧ refs .first = st[r2 -- fs '] .refval ∧ validrefs(str, refs, st[st[r2 -- fs '] .refval -- _type] .type, st), r0 ≠ r2 ∧ r0 ≠ st[r2 -- fs '] .refval ∧ ¬ r0 ∈ refs

User applied rule cut formula

And obtained the following new goals

Rule argument

[fma-arg: r0 = r2]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (r0 = r2)]]