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Proof step 15 from proof add-in-fskey (previous step)

is_fskey(sk), r -- sk ∈ addarray(r, ty, vals, st), ∀ st, ty, sk, r, vals0. # vals0 < # vals → (r -- sk ∈ addarray(r, ty, vals0, st) ∧ is_fskey(sk) → sk = _length ∨ sk = _type ∨ r -- sk ∈ st) ⊦ sk = _type, sk = _length, r -- sk ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ vals ≠ [] → addarray(r, ty, vals, st) = (addarray(r, ty, vals .butlast, st)[r -- n→i((# vals) -1) ', vals .last][r -- _length, intval(n→i(# vals))])
Proof step 8:
¬ vals ≠ [], is_fskey(sk), r -- sk ∈ addarray(r, ty, vals, st), ∀ st, ty, sk, r, vals0. # vals0 < # vals → (r -- sk ∈ addarray(r, ty, vals0, st) ∧ is_fskey(sk) → sk = _length ∨ sk = _type ∨ r -- sk ∈ st) ⊦ sk = _type, sk = _length, r -- sk ∈ st
Proof step 14:
vals ≠ [], is_fskey(sk), r -- sk ∈ (addarray(r, ty, vals .butlast, st)[r -- n→i((# vals) -1) ', vals .last][r -- _length, intval(n→i(# vals))]), ∀ st, ty, sk, r, vals0. # vals0 < # vals → (r -- sk ∈ addarray(r, ty, vals0, st) ∧ is_fskey(sk) → sk = _length ∨ sk = _type ∨ r -- sk ∈ st) ⊦ sk = _type, sk = _length, r -- sk ∈ st

Rule argument

[crewrite-arg: add-lastrec, F, [st, vals, ty, r] → [st, vals, ty, r]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, vals, ty, r)]]