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Proof step 17 from proof addbar-refval-01 (previous step)

# 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 '])) ⊦ ints = [], is2bs(ints) = [], is_bytevalue((i = n→i(m) ⊃ mkjavavalues(is2bs(ints)) .last; addarray(r, byte_type, mkjavavalues(is2bs(ints)) .butlast, st)[r -- i ']))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 9:
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(mkjavavalues(is2bs(ints)) .last), ints = [], is2bs(ints) = []
Proof step 16:
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) = []

Rule argument

[CASEDARG : pos: right 3, ()]

Additional information

[[CONCRETESIMPRULES : ()]]