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Proof step 6 from proof ints2bytes-at (previous step)

n ≠ 0, n ≤ # ints0, ∀ n0. ¬ # ints0 ≤ n0 → (is2bs(ints0)[n0]) = i2b(ints0[n0]) ∨ ints0 = [] ⊦ (is2bs(ints0)[n -1]) = i2b((i ' + ints0)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n ≤ # ints0, ∀ n0. ¬ # ints0 ≤ n0 → (is2bs(ints0)[n0]) = i2b(ints0[n0]) ∨ ints0 = [] ⊦ n = 0, (is2bs(ints0)[n - 1]) = i2b((i ' + ints0)[n])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ n -1 = n - 1])]]