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Proof step 18 from proof int2bytes-byte (previous step)

i + 2 ^ (6 + n→i(n0)) < 0, 0 ≤ 2 ^ (7 + n→i(n0)) + i, i < 0, ~128 ≤ i ⊦ bits2bytes(firstn(n0, int2bits(i))) + i2b(bits2int(lastn(8, int2bits(i)))) ' = i2b(i) ', n0 = 0

User applied rule cut formula

And obtained the following new goals

Proof step 19:
~(2 ^ 7) + 2 ^ (6 + n→i(n0)) < 0, i + 2 ^ (6 + n→i(n0)) < 0, 0 ≤ 2 ^ (7 + n→i(n0)) + i, i < 0, ~128 ≤ i ⊦ bits2bytes(firstn(n0, int2bits(i))) + i2b(bits2int(lastn(8, int2bits(i)))) ' = i2b(i) ', n0 = 0
Proof step 23:
i + 2 ^ (6 + n→i(n0)) < 0, 0 ≤ 2 ^ (7 + n→i(n0)) + i, i < 0, ~128 ≤ i ⊦ ~(2 ^ 7) + 2 ^ (6 + n→i(n0)) < 0, bits2bytes(firstn(n0, int2bits(i))) + i2b(bits2int(lastn(8, int2bits(i)))) ' = i2b(i) ', n0 = 0

Rule argument

fma: ~(2 ^ 7) + 2 ^ (6 + n→i(n0)) < 0

Additional information

Used formulas: ~(2 ^ 7) + 2 ^ (6 + n→i(n0)) < 0