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Proof step 16 from proof addbar-refval-01 (previous step)

i ≠ n→i(m), # ints = m + 1, i ≤ n→i(m), 0 ≤ i, ∀ st, r, i, ints0. # ints0 < m + 1 → (0 ≤ i ∧ i < n→i(# ints0) → ints0 = [] ∨ is_bytevalue(addarray(r, byte_type, mkjavavalues(is2bs(ints0)), st)[r -- i '])) ⊦ is_bytevalue(addarray(r, byte_type, mkjavavalues(is2bs(ints)) .butlast, st)[r -- i ']), ints = [], is2bs(ints) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# ints = m + 1, i < n→i(m), 0 ≤ i, ∀ st, r, i, ints0. # ints0 < m + 1 → (0 ≤ i ∧ i < n→i(# ints0) → ints0 = [] ∨ is_bytevalue(addarray(r, byte_type, mkjavavalues(is2bs(ints0)), st)[r -- i '])) ⊦ i = n→i(m), is2bs(ints) = [], ints = [], m = 0, is_bytevalue(addarray(r, byte_type, mkjavavalues(is2bs(ints)) .butlast, st)[r -- i '])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 < n ↔ n ≠ 0], [CSIMPSEQ : 0 ≤ i ⊦ ((* i)) < n→i(n) ↔ i→n(i) < n], [CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPSEQ : ⊦ i ≠ j → (i ≤ j ↔ i < j)])]]