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Proof step 28 from proof int2bytes-byte (previous step)

2 ^ (6 + n→i(n0)) < 128, 0 ≤ i, i < 2 ^ (7 + n→i(n0)), 2 ^ (6 + n→i(n0)) ≤ i, i < 128 ⊦ 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 29:
128 = 2 ^ 7, 2 ^ (6 + n→i(n0)) < 128, 0 ≤ i, i < 2 ^ (7 + n→i(n0)), 2 ^ (6 + n→i(n0)) ≤ i, i < 128 ⊦ bits2bytes(firstn(n0, int2bits(i))) + i2b(bits2int(lastn(8, int2bits(i)))) ' = i2b(i) ', n0 = 0

Rule argument

fma: 128 = 2 ^ 7

Additional information

Used formulas: 128 = 2 ^ 7