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Proof step 9 from proof add-isobj (previous step)

∀ st, ty, sks, class, r0, r, vals0. # vals0 < # vals → (true → r = r0 ∨ (is_obj(r0, class, sks, addarray(r, ty, vals0, st)) ↔ is_obj(r0, class, sks, st))) ⊦ is_obj(r0, class, sks, addarray(r, ty, vals, st)) ↔ is_obj(r0, class, sks, st), r = r0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ 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 6:
¬ vals ≠ [], ∀ st, ty, sks, class, r0, r, vals0. # vals0 < # vals → (true → r = r0 ∨ (is_obj(r0, class, sks, addarray(r, ty, vals0, st)) ↔ is_obj(r0, class, sks, st))) ⊦ is_obj(r0, class, sks, addarray(r, ty, vals, st)) ↔ is_obj(r0, class, sks, st), r = r0
Proof step 8:
vals ≠ [], ∀ st, ty, sks, class, r0, r, vals0. # vals0 < # vals → (true → r = r0 ∨ (is_obj(r0, class, sks, addarray(r, ty, vals0, st)) ↔ is_obj(r0, class, sks, st))) ⊦ is_obj(r0, class, sks, addarray(r, ty, vals .butlast, st)[r -- n→i((# vals) -1) ', vals .last][r -- _length, intval(n→i(# vals))]) ↔ is_obj(r0, class, sks, st), r = r0

Rule argument

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

Additional information

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