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Proof step 53 from proof hassharing-validrefs-dups (previous step)

st[r -- i '] = refval(r1), ∃ sks1, sks2. fits(r1, sks1, st) ∧ fits(r1, sks2, st) ∧ sks1 ≠ sks2 ∧ sks1 ≠ [] ∧ sks2 ≠ [] ∧ st[r1 -- sks1] .refval ≠ jvmref ∧ st[r1 -- sks1] = st[r1 -- sks2], validrefs(str, r1 ' + refs1, st[r1 -- _type] .type, st), is_array(str, r, st), i < st[r -- _length] .intval, validrefs(str, r1 ' + refs1, st[r -- _type] .type .type, st), 0 ≤ i, 0 < st[r -- _length] .intval, ∀ st0, str, refs, r0. # (r0 ' + refs) < # (r ' + (r1 + refs1)) → (validrefs(str, r0 ' + refs, st0[r0 -- _type] .type, st0) → hasSharing(r0, st0) ∨ ¬ r0 ∈ refs ∧ ¬ dups(refs)) ⊦ r = jvmref, ∃ sks1, sks2. fits(r, sks1, st) ∧ fits(r, sks2, st) ∧ sks1 ≠ sks2 ∧ sks1 ≠ [] ∧ sks2 ≠ [] ∧ st[r -- sks1] .refval ≠ jvmref ∧ st[r -- sks1] = st[r -- sks2], r ∈ refs1, ¬ r1 ∈ refs1 ∧ ¬ dups(refs1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 52:
st[r -- i '] = refval(r1), st[r1 -- sks0] = st[r1 -- sks], fits(r1, sks0, st), fits(r1, sks, st), validrefs(str, r1 ' + refs1, st[r1 -- _type] .type, st), is_array(str, r, st), i < st[r -- _length] .intval, validrefs(str, r1 ' + refs1, st[r -- _type] .type .type, st), 0 ≤ i, 0 < st[r -- _length] .intval, ∀ st0, str, refs, r0. # (r0 ' + refs) < # (r ' + (r1 + refs1)) → (validrefs(str, r0 ' + refs, st0[r0 -- _type] .type, st0) → hasSharing(r0, st0) ∨ ¬ r0 ∈ refs ∧ ¬ dups(refs)) ⊦ r = jvmref, st[r1 -- sks] .refval = jvmref, sks = [], sks0 = [], sks0 = sks, r ∈ refs1, ∃ sks1, sks2. fits(r, sks1, st) ∧ fits(r, sks2, st) ∧ sks1 ≠ sks2 ∧ sks1 ≠ [] ∧ sks2 ≠ [] ∧ st[r -- sks1] .refval ≠ jvmref ∧ st[r -- sks1] = st[r -- sks2], ¬ r1 ∈ refs1 ∧ ¬ dups(refs1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]