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Proof step 5 from proof int2bytes-len-eq-pos (previous step)

int2bytes(i).first = i2b(0), 0 ≤ i, ¬ 0 ≤ b2i(int2bytes(i).rest.first) → bytes2int(i2b(0) ' + int2bytes(i).rest.first ' + int2bytes(i).rest.rest) < 2 ^ ((n→i(# int2bytes(i).rest.rest) + 1) * 8) ⊦ # int2bytes(i) = n ↔ ¬ i < 2 ^ ((n→i(n) - 1) * 8 - 1) ∧ ¬ 2 ^ ((n→i(n) - 1) * 8) ≤ i ∧ n ≠ 0

User applied rule case distinction

And obtained the following new goals

Proof step 6:
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
Proof step 10:
bytes2int(i2b(0) ' + int2bytes(i).rest.first ' + int2bytes(i).rest.rest) < 2 ^ ((n→i(# int2bytes(i).rest.rest) + 1) * 8), int2bytes(i).first = i2b(0), 0 ≤ i ⊦ # int2bytes(i) = n ↔ ¬ i < 2 ^ ((n→i(n) - 1) * 8 - 1) ∧ ¬ 2 ^ ((n→i(n) - 1) * 8) ≤ i ∧ n ≠ 0

Rule argument

pos: left 3

Proof step 5 from proof int2bytes-len-eq-pos (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information