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Proof step 24 from proof bytes2int-diff (previous step)

bytes2int(bytes2) * 256 + band(b2i(by), 255) = bytes2int(bytes3) * 256 + band(b2i(by0), 255), # bytes2 = # bytes3, ∀ bytes0, bytes1. # bytes1 < #(bytes3 + by0 ') → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ bytes2 + by ' = bytes3 + by0 ', bytes3 = []

User applied rule cut formula

And obtained the following new goals

Proof step 25:
by = by0, bytes2int(bytes2) * 256 + band(b2i(by), 255) = bytes2int(bytes3) * 256 + band(b2i(by0), 255), # bytes2 = # bytes3, ∀ bytes0, bytes1. # bytes1 < #(bytes3 + by0 ') → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ bytes2 + by ' = bytes3 + by0 ', bytes3 = []
Proof step 27:
bytes2int(bytes2) * 256 + band(b2i(by), 255) = bytes2int(bytes3) * 256 + band(b2i(by0), 255), # bytes2 = # bytes3, ∀ bytes0, bytes1. # bytes1 < #(bytes3 + by0 ') → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ by = by0, bytes2 + by ' = bytes3 + by0 ', bytes3 = []

Rule argument

fma: by = by0

Additional information

Used formulas: by = by0