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Proof step 103 from proof pseq-subset (previous step)

st2[r -- _type] = st1[r1 -- _type], st[r -- _type] = st1[r1 -- _type], st1[r1 -- _type] .type = mkclasstype(class), r -- _type ∈ st, r -- _type ∈ st2, r ∈ st, r ∈ st2, ∀ fs. fs ' ∈ instfields(class) ∧ ¬ is_referencevalue(st[r -- fs ']) → st[r -- fs '] = st1[r1 -- fs '], ∀ fs. fs ' ∈ instfields(class) ∧ is_referencevalue(st[r -- fs ']) → ps=(st[r -- fs '] .refval, st, st1[r1 -- fs '] .refval, st1), ∀ rk. rk ∈ st → rk ∈ st2 ∧ st[rk] = st2[rk], ∀ 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)) ⊦ r1 = jvmref, r = jvmref, cyclic(str, r1, st1), junkref(str, r1, st1), cyclic(str, r, st), junkref(str, r, st), (fs ' ∈ instfields(class) ∧ is_referencevalue(st2[r -- fs ']) → ps=(st2[r -- fs '] .refval, st2, st1[r1 -- fs '] .refval, st1)) ∧ (fs0 ' ∈ instfields(class) ∧ ¬ is_referencevalue(st2[r -- fs0 ']) → st2[r -- fs0 '] = st1[r1 -- fs0 '])

User applied rule case distinction

And obtained the following new goals

Rule argument

[CASEDARG : pos: right 7, ()]

Additional information

[[CONCRETESIMPRULES : ()]]