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Proof step 20 from proof fits-append (previous step)

∀ r, sks0, st. fits(r, sks1 + sks0, st) ↔ fits(r, sks1, st) ∧ fits(st[r -- sks1] .refval, sks0, st) ⊦ fits(r, (sk ' + sks1) + sks0, st) ↔ fits(r, sk ' + sks1, st) ∧ fits(st[r -- (sk ' + sks1)] .refval, sks0, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ sks ≠ [] → st[r -- (sk ' + sks)] = st[st[r -- sk] .refval -- sks]
Proof step 9:
¬ sks1 ≠ [], ∀ r, sks0, st. fits(r, sks1 + sks0, st) ↔ fits(r, sks1, st) ∧ fits(st[r -- sks1] .refval, sks0, st) ⊦ fits(r, (sk ' + sks1) + sks0, st) ↔ fits(r, sk ' + sks1, st) ∧ fits(st[r -- (sk ' + sks1)] .refval, sks0, st)
Proof step 19:
sks1 ≠ [], ∀ r, sks0, st. fits(r, sks1 + sks0, st) ↔ fits(r, sks1, st) ∧ fits(st[r -- sks1] .refval, sks0, st) ⊦ fits(r, (sk ' + sks1) + sks0, st) ↔ fits(r, sk ' + sks1, st) ∧ fits(st[st[r -- sk] .refval -- sks1] .refval, sks0, st)

Rule argument

[speclemmaarg: ,sks ≠ [] → st[r -- (sk ' + sks)] = st[st[r -- sk] .refval -- sks] ,[sks, sk, r, st] → [sks1, sk, r, st] ,F ,storefuns , ,lookup-rec-01 ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sks1, sk, r, st)]]