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Proof step 93 from proof bytes2int-int2bytes-len-eq (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 94:
⊦ i2b(i) ≠ i2b(j) → i ≠ j
Proof step 95:
¬ i2b(b2i(by)) ≠ i2b(0), b2i(by) = 0, # bytes1 = # int2bytes(bytes2int(bytes1)), bytes2int(bytes1) < 0, bytes2int(bytes1) < ~1, ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes1 = [], bytes1 + by ' = int2bytes(bytes2int(bytes1)) + i2b(0) ', bytes2int(bytes1) * 256 = ~256

Rule argument

seq: ⊦ i2b(i) ≠ i2b(j) → i ≠ j subst: [j, i] → [0, b2i(by)] rotate: F spec: byte inst: name: i2b-noteq

Additional information

Used terms: 0, b2i(by)