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Proof step 30 from proof lookup-put-ex-first (previous step)

sks ≠ [], ∀ sks2. sks2 + sk ' ⊑ sks → st[st[r -- sk0] .refval -- sks2] .refval ≠ st[st[r -- sk0] .refval -- sks] .refval, r ≠ st[st[r -- sk] .refval -- sks] .refval ∨ sk0 ≠ sk ⊦ (st[st[st[r -- sk0] .refval -- sks] .refval -- sk, val][st[r -- sk0] .refval -- (sks + sk ' + sks1)]) = st[st[r -- sk0] .refval -- (sks + sk ' + sks1)], ∃ sks2, sks3. sk0 ' + sks + sk ' + sks1 = sks2 + sk ' + sks3 ∧ st[r -- sks2] .refval = st[st[r -- sk0] .refval -- sks] .refval ∧ (∀ sks3. sks3 + sk ' ⊑ sks2 → st[r -- sks3] .refval ≠ st[r -- sks2] .refval), st[st[r -- sk0] .refval -- sks] .refval = st[st[r -- sk0] .refval -- sks] .refval ∧ (sks0 + sk ' ⊑ sk0 ' + sks → st[r -- sks0] .refval ≠ st[st[r -- sk0] .refval -- sks] .refval)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
st[r -- sks0] .refval = st[st[r -- sk0] .refval -- sks] .refval, sks0 + sk ' ⊑ sk0 ' + sks, ∀ sks2. sks2 + sk ' ⊑ sks → st[st[r -- sk0] .refval -- sks2] .refval ≠ st[r -- sks0] .refval, r ≠ st[st[r -- sk] .refval -- sks] .refval ∨ sk0 ≠ sk ⊦ (st[st[r -- sks0] .refval -- sk, val][st[r -- sk0] .refval -- (sks + sk ' + sks1)]) = st[st[r -- sk0] .refval -- (sks + sk ' + sks1)], sks = [], ∃ sks2, sks3. sk0 ' + sks + sk ' + sks1 = sks2 + sk ' + sks3 ∧ st[r -- sks2] .refval = st[r -- sks0] .refval ∧ (∀ sks3. sks3 + sk ' ⊑ sks2 → st[r -- sks3] .refval ≠ st[r -- sks2] .refval)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]