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Proof step 14 from proof putarray-add-same-02 (previous step)

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 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
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)) ⊦ vals = [], putArray(vals, r, 0, addarray(r, initial_value(ty), i, st)[r -- i ', initial_value(ty)]) = (addarray(r, vals .butlast, 0, st)[r -- i ', vals .last])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ((* i)) + i0 - ((* j)) = (i - j) + i0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + 0 = i])]]