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Proof step 7 from proof add-refs (previous step)

vals ≠ [], ∀ st, ty, r, vals0. # vals0 < # vals → (true → addarray(r, ty, vals0, st) .refs = st .refs ++ r) ⊦ (addarray(r, ty, vals .butlast, st)[r -- n→i((# vals) -1) ', vals .last][r -- _length, intval(n→i(# vals))]) .refs = st .refs ++ r

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ st, ty, r, vals0. # vals0 < # vals → (true → addarray(r, ty, vals0, st) .refs = st .refs ++ r) ⊦ vals = [], addarray(r, ty, vals .butlast, st) .refs ++ r = st .refs ++ r

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rset ++ r ++ r = rset ++ r], [CSIMPSEQ : ⊦ (r -- sk) .ref = r], [CSIMPSEQ : ⊦ st[rk, val] .refs = st .refs ++ rk .ref], [CSIMPSEQ : ⊦ n -1 = n - 1])]]