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Proof step 45 from proof fits-in (previous step)

fits(r, sk ' + sks0, st), ∀ str, r, st. sks0 ≠ [] ∧ ¬ junkref(str, r, st) ∧ fits(r, sks0, st) → r -- sks0 ∈ st ⊦ junkref(str, r, st), r -- (sk ' + sks0) ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ sks ≠ [] → (r -- (sk ' + sks) ∈ st ↔ r -- sk ∈ st ∧ st[r -- sk] .refval -- sks ∈ st)
Proof step 4:
¬ sks0 ≠ [], fits(r, sk ' + sks0, st), ∀ str, r, st. sks0 ≠ [] ∧ ¬ junkref(str, r, st) ∧ fits(r, sks0, st) → r -- sks0 ∈ st ⊦ junkref(str, r, st), r -- (sk ' + sks0) ∈ st
Proof step 44:
sks0 ≠ [], fits(r, sk ' + sks0, st), ∀ str, r, st. sks0 ≠ [] ∧ ¬ junkref(str, r, st) ∧ fits(r, sks0, st) → r -- sks0 ∈ st ⊦ junkref(str, r, st), r -- sk ∈ st ∧ st[r -- sk] .refval -- sks0 ∈ st

Rule argument

[speclemmaarg: , sks ≠ [] → ( r -- (sk ' + sks) ∈ st ↔ r -- sk ∈ st ∧ st[r -- sk] .refval -- sks ∈ st) ,[st, sks, sk, r] → [st, sks0, sk, r] ,F ,storefuns , ,instore-rec-01 ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, sks0, sk, r)]]