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Proof step 15 from proof putarray-add-same-02 (previous step)

n→i(# vals) = i + 1, ∀ st, ty, r, vals0. # vals0 < # vals → (true → putArray(vals0, r, 0, addarray(r, initial_value(ty), n→i(# vals0), st)) = addarray(r, vals0, 0, st)) ⊦ putArray(vals, r, 0, addarray(r, initial_value(ty), i, st)[r -- i ', initial_value(ty)]) = addarray(r, vals, 0, st), vals = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ vals ≠ [] → addarray(r, vals, i, st) = (addarray(r, vals .butlast, i, st)[r -- (i + n→i(# vals) - 1) ', vals .last])
Proof step 14:
vals ≠ [], n→i(# vals) = i + 1, ∀ st, ty, r, vals0. # vals0 < # vals → (true → putArray(vals0, r, 0, addarray(r, initial_value(ty), n→i(# vals0), st)) = addarray(r, vals0, 0, st)) ⊦ putArray(vals, r, 0, addarray(r, initial_value(ty), i, st)[r -- i ', initial_value(ty)]) = (addarray(r, vals .butlast, 0, st)[r -- (0 + n→i(# vals) - 1) ', vals .last]), vals = []

Rule argument

[crewrite-arg: addarray-lastrec , F , [st, i, vals, r] → [st, 0, vals, r] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, 0, vals, r)]]