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Proof step 106 from proof bytes2int-int2bytes-len-eq (previous step)

256 * b2i(by0) + band(b2i(by), 255) = 0 - 256, ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ by0 ' + by ' = i2b(~1) ' + i2b(0) '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 107:
⊦ i0 = i - j ↔ i = i0 + j
Proof step 108:
0 = (256 * b2i(by0) + band(b2i(by), 255)) + 256, ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ by0 ' + by ' = i2b(~1) ' + i2b(0) '

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [j, i, i0] → [256, 0, 256 * b2i(by0) + band(b2i(by), 255)] rotate: F spec: int-abs inst: name: elim-sub

Additional information

Used terms: 256, 0, 256 * b2i(by0) + band(b2i(by), 255)