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Proof step 8 from proof ints2bytes-at (previous step)

∀ n. n < # ints0 → (is2bs(ints0)[n]) = i2b(ints0[n]) ∨ ints0 = [], n < # (i + ints0) ⊦ (is2bs(i + ints0)[n]) = i2b((i + ints0)[n]), i + ints0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ is2bs(i ' + ints) = i2b(i) ' + is2bs(ints)
Proof step 7:
n ≤ # ints0, ∀ n0. ¬ # ints0 ≤ n0 → (is2bs(ints0)[n0]) = i2b(ints0[n0]) ∨ ints0 = [] ⊦ (i2b(i) ' + is2bs(ints0))[n] = i2b((i ' + ints0)[n])

Rule argument

[emptyarg]

Additional information

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