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Proof step 23 from proof getbar-addbar-02 (previous step)

n < # sublist(i→n(i), i→n(j), ints), n < i→n(j), i + j ≤ n→i(# ints), i→n(i) + i→n(j) < # ints, 0 ≤ i, 0 < j ⊦ (mkjavavalues(is2bs(ints))[i→n(i) + n]) .byteval = i2b(sublist(i→n(i), i→n(j), ints)[n]), ints = [], j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
0 < j, 0 ≤ i, i→n(i) + i→n(j) < # ints, i + j ≤ n→i(# ints), n < i→n(j), n < # sublist(i→n(i), i→n(j), ints) ⊦ j = 0, ints = [], (mkjavavalues(is2bs(ints))[i→n(i) + n]) .byteval = i2b(sublist(i→n(i), i→n(j), ints)[n]), sublist(i→n(i), i→n(j), ints) = []

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # ints = 0 ↔ ints = []], [CSIMPFORWARD : ⊦ m < n → n ≠ 0])]]