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Proof step 44 from proof lookup-put-ex-first (previous step)

∀ st, r0, sk, val, r. (st[r0 -- sk, val][r -- sks0]) ≠ st[r -- sks0] → (∃ sks2, sks1. sks0 = sks2 + (sk + sks1) ∧ st[r -- sks2] .refval = r0 ∧ (∀ sks3. sks3 + sk ⊑ sks2 → st[r -- sks3] .refval ≠ r0)), (st[r0 -- sk, val][r -- (sk0 + sks0)]) ≠ st[r -- (sk0 + sks0)] ⊦ ∃ sks2, sks1. sk0 + sks0 = sks2 + (sk + sks1) ∧ st[r -- sks2] .refval = r0 ∧ (∀ sks3. sks3 + sk ⊑ sks2 → st[r -- sks3] .refval ≠ r0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
∀ st0, r1, sk1, val0, r2. (st0[r1 -- sk1, val0][r2 -- sks0]) ≠ st0[r2 -- sks0] → (∃ sks, sks1. sks0 = sks + sk1 ' + sks1 ∧ st0[r2 -- sks] .refval = r1 ∧ (∀ sks2. sks2 + sk1 ' ⊑ sks → st0[r2 -- sks2] .refval ≠ st0[r2 -- sks] .refval)) ⊦ (st[r0 -- sk, val][r -- (sk0 ' + sks0)]) = st[r -- (sk0 ' + sks0)], ∃ sks, sks1. sk0 ' + sks0 = sks + sk ' + sks1 ∧ st[r -- sks] .refval = r0 ∧ (∀ sks2. sks2 + sk ' ⊑ sks → st[r -- sks2] .refval ≠ st[r -- sks] .refval)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, storekeys, Last, ⊦ sks + sk = sks + sk '], [CSIMPNAMEDSEQ : list, storekeys, c, ⊦ sk + sks = sk ' + sks])]]