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Proof step 161 from proof bytes2int-int2bytes-len-eq (previous step)

(b2i(by0) - 1) * 256 + band(b2i(by), 255) < 0, ~129 < (b2i(by0) - 1) * 256 + band(b2i(by), 255), 0 ≤ b2i(by0), ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ by0 = i2b(0), by = i2b(0), ¬ b2i(by0) ≤ 1 ∧ ¬ 256 ≤ abs(band(b2i(by), 255))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 162:
b2i(by0) = i + 1, i * 256 + band(b2i(by), 255) < 0, ~129 < i * 256 + band(b2i(by), 255), 0 ≤ b2i(by0), ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ by0 = i2b(0), by = i2b(0), ¬ b2i(by0) ≤ 1 ∧ ¬ 256 ≤ abs(band(b2i(by), 255))

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [b2i(by0), 1, i] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: b2i(by0), 1, i