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Proof step 131 from proof pseq-subset (previous step)

st2[r -- i '] = st[r -- i '], st[r -- _type] = st1[r1 -- _type], st2[r -- _type] = st1[r1 -- _type], st[r -- _length] .intval = st1[r1 -- _length] .intval, st2[r -- _length] .intval = st1[r1 -- _length] .intval, 0 < st1[r1 -- _length] .intval, 0 ≤ i, is_arraytype(st1[r1 -- _type] .type), r ∈ st2, r ∈ st, r -- _type ∈ st2, r -- _type ∈ st, is_array(str, r1, st1), is_array(str, r, st), is_referencevalue(st[r -- i ']), i < st1[r1 -- _length] .intval, r -- i ' ∈ st2, r -- i ' ∈ st, ps=(st[r -- i '] .refval, st, jvmref, st1), ∀ i. ¬ i < 0 ∧ ¬ st1[r1 -- _length] .intval ≤ i ∧ is_referencevalue(st[r -- i ']) → ps=(st[r -- i '] .refval, st, st1[r1 -- i '] .refval, st1), ∀ i. ¬ i < 0 ∧ ¬ st1[r1 -- _length] .intval ≤ i ∧ ¬ is_referencevalue(st[r -- i ']) → st[r -- i '] = st1[r1 -- i '], ∀ rk. rk ∈ st → rk ∈ st2 ∧ st[rk] = st2[rk], junkref(str, st[r -- i '] .refval, st) ∨ cyclic(str, st[r -- i '] .refval, st), ∀ 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, is_referencevalue(st1[r1 -- i ']), cyclic(str, r, st), cyclic(str, r1, st1), ps=(st[r -- i '] .refval, st2, jvmref, st1), is_reftype(st1[r1 -- _type] .type .type), st[r -- i '] .refval ≠ jvmref ∧ (st[st[r -- i '] .refval -- _type] .type ≠ st1[r1 -- _type] .type .type ∨ junkref(str, st[r -- i '] .refval, st))

User applied rule cut formula

And obtained the following new goals

Rule argument

[fma-arg: st[r -- i '] .refval ≠ jvmref]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (st[r -- i '] .refval ≠ jvmref)]]