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Proof step 7 from proof int2bytes-len-eq-pos (previous step)

int2bytes(i).first = i2b(0), 0 ≤ i, 0 ≤ b2i(int2bytes(i).rest.first) ⊦ # int2bytes(i) = n ↔ ¬ i < 2 ^ ((n→i(n) - 1) * 8 - 1) ∧ ¬ 2 ^ ((n→i(n) - 1) * 8) ≤ i ∧ n ≠ 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
int2bytes(i) = by + bytes ∧ int2bytes(i) ≠ [], by = i2b(0), 0 ≤ i, 0 ≤ b2i(bytes.first) ⊦ # int2bytes(i) = n ↔ ¬ i < 2 ^ ((n→i(n) - 1) * 8 - 1) ∧ ¬ 2 ^ ((n→i(n) - 1) * 8) ≤ i ∧ n ≠ 0

Rule argument

seq: ⊦ bytes ≠ [] → (by = bytes.first ∧ bytes0 = bytes.rest ↔ bytes = by + bytes0) subst: [bytes, by, bytes0] → [int2bytes(i), by, bytes] rotate: T spec: list-data inst: bytelist name: elim-1

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];%⊦ bytes ≠ [] → (by = bytes.first ∧ bytes0 = bytes.rest ↔ bytes = by + bytes0); (elim) Used terms: int2bytes(i), by, bytes