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Proof step 17 from proof addarray-both-subset (previous step)

st0 ⊆ st1, ∀ r, st1, st0, vals0. # vals0 < # vals → (st0 ⊆ st1 → addarray(r, byte_type, vals0, st0) ⊆ addarray(r, byte_type, vals0, st1)) ⊦ vals = [], addarray(r, byte_type, vals, st0) ⊆ addarray(r, byte_type, vals, st1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ 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 16:
vals ≠ [], st0 ⊆ st1, ∀ r, st1, st0, vals0. # vals0 < # vals → (st0 ⊆ st1 → addarray(r, byte_type, vals0, st0) ⊆ addarray(r, byte_type, vals0, st1)) ⊦ vals = [], (addarray(r, byte_type, vals .butlast, st0)[r -- n→i((# vals) -1) ', vals .last][r -- _length, intval(n→i(# vals))]) ⊆ addarray(r, byte_type, vals, st1)

Rule argument

[crewrite-arg: add-lastrec , F , [st, vals, ty, r] → [st0, vals, byte_type, r] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0, vals, byte_type, r)]]