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Proof step 134 from proof bytes2int-int2bytes-len-eq (previous step)

1 = # bytes1, 0 < bytes2int(bytes1) * 256 + band(b2i(by), 255), bytes2int(bytes1) * 256 + band(b2i(by), 255) ≤ 255, 127 < bytes2int(bytes1) * 256 + band(b2i(by), 255), ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes1 = [], bytes1 + by ' = i2b(0) ' + i2b(bytes2int(bytes1) * 256 + band(b2i(by), 255) - 256) '

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 135:
bytes2int(bytes1) * 256 + band(b2i(by), 255) = i + 256, 1 = # bytes1, 0 < bytes2int(bytes1) * 256 + band(b2i(by), 255), bytes2int(bytes1) * 256 + band(b2i(by), 255) ≤ 255, 127 < bytes2int(bytes1) * 256 + band(b2i(by), 255), ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes1 = [], bytes1 + by ' = i2b(0) ' + i2b(i) '

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [bytes2int(bytes1) * 256 + band(b2i(by), 255), 256, i] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: bytes2int(bytes1) * 256 + band(b2i(by), 255), 256, i