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Proof step 36 from proof lookup-put-ex-first (previous step)

r -- sk0 ≠ r0 -- sk, ∀ 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)), r ≠ r0 ∨ sk0 ≠ sk ⊦ sks0 = [], (st[r0 -- sk, val][st[r -- sk0] .refval -- sks0]) = st[st[r -- sk0] .refval -- 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
∀ 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)), r ≠ r0 ∨ sk0 ≠ sk ⊦ sks0 = [], (st[r0 -- sk, val][st[r -- sk0] .refval -- sks0]) = st[st[r -- sk0] .refval -- 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 : refkey, , inj-01, ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)])]]