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Proof step 4 from proof ints2bytes-at (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m < # ints0, ∀ n. ¬ # ints0 ≤ n → (is2bs(ints0)[n]) = i2b(ints0[n]) ⊦ (is2bs(ints0)[m]) = i2b(ints0[m]), ints0 = []

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # ints = 0 ↔ ints = []], [CSIMPFORWARD : ⊦ m < n → n ≠ 0], [CSIMPSEQ : ⊦ m + 1 ≤ n ↔ m < n], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ (i ' + ints)[n + 1] = ints[n]], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + 1 ≠ 0])]]