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Proof step 6 from proof put-last-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ st, r, sk, val. st[r -- (sks0 + sk '), val] = st[st[r -- sks0] .refval -- sk, val] ⊦ sks0 = [], st[st[r -- sk0] .refval -- (sks0 + sk '), val] = st[st[st[r -- sk0] .refval -- sks0] .refval -- sk, val]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]