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Proof step 42 from proof bytes2int-diff (previous step)

b2i(by) < 128, b2i(by) = 256 + b2i(by0), # bytes2 = # bytes3, bytes2int(bytes2) = bytes2int(bytes3), b2i(by0) < 0, 0 ≤ b2i(by), ∀ bytes0, bytes1. # bytes1 < #(bytes3 + by0 ') → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ by = by0, bytes2 + by ' = bytes3 + by0 ', bytes3 = []

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 2 flags: discard

Additional information

Used simplifier rules: ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ ¬ b2i(by) < ~128;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;