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Proof step 9 from proof at-put-same (previous step)

sks0 ≠ [], ∀ st0, r0, val0. (st0[r0 -- sks0, val0][r0 -- sks0]) = val0 ⊦ sks0 = [], (st[st[r -- sk] .refval -- sks0, val][(st[st[r -- sk] .refval -- sks0, val][r -- sk]) .refval -- sks0]) = val, sk ∈ sks0, dups(sks0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ st0, r0, val0. (st0[r0 -- sks0, val0][r0 -- sks0]) = val0 ⊦ sks0 = [], (st[st[r -- sk] .refval -- sks0, val][(st[st[r -- sk] .refval -- sks0, val][r -- sk]) .refval -- sks0]) = val, dups(sks0), sk ∈ sks0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]