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Proof step 5 from proof ints2bytes-at (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
n = m + 1, n ≤ # ints0, ∀ n0. ¬ # ints0 ≤ n0 → (is2bs(ints0)[n0]) = i2b(ints0[n0]) ∨ ints0 = [] ⊦ n = 0, (is2bs(ints0)[m]) = i2b((i ' + ints0)[n])

Rule argument

[speclemmaarg: n ≠ 0 ,m = n - 1 ↔ n = m + 1 ,[n, m] → [n, m] ,T ,nat , ,elim-pred-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1])], [EXTRATERMS : (n, m)]]