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Proof step 11 from proof hassharing-validrefs-dups (previous step)

st[r -- sks] = refval(r1) ∧ is_referencevalue(st[r -- sks]), # sks = 1 + # refs1, r1 = r, validrefs(str, r ' + refs1 + r ', st[r -- _type] .type, st), fits(r, sks, st), is_valid_ref(str, r, st), ∀ n. n ≠ 0 ∧ ¬ # sks < n → st[r -- firstn(n, sks)] .refval = (r ' + refs1 + r ')[n], ∀ st, str, refs2, r0. # (r0 ' + refs2) < # (r ' + refs1 + (r + refs)) → (validrefs(str, r0 ' + refs2, st[r0 -- _type] .type, st) → hasSharing(r0, st) ∨ ¬ r0 ∈ refs2 ∧ ¬ dups(refs2)) ⊦ sks = [], r = jvmref, st[r -- sks] = st[r -- (sks + sks)] ∧ fits(r, sks + sks, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# sks = 1 + # refs1, st[r -- sks] = refval(r), is_valid_ref(str, r, st), fits(r, sks, st), validrefs(str, r ' + refs1 + r ', st[r -- _type] .type, st), ∀ n. n ≠ 0 ∧ ¬ # sks < n → st[r -- firstn(n, sks)] .refval = (r ' + refs1 + r ')[n], ∀ st, str, refs2, r0. # (r0 ' + refs2) < # (r ' + refs1 + (r + refs)) → (validrefs(str, r0 ' + refs2, st[r0 -- _type] .type, st) → hasSharing(r0, st) ∨ ¬ r0 ∈ refs2 ∧ ¬ dups(refs2)) ⊦ r = jvmref, sks = [], refval(r) = st[r -- (sks + sks)] ∧ fits(r, sks + sks, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ is_referencevalue(refval(r))])]]