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Proof step 23 from proof bytes2int-diff (previous step)

bytes = bytes3 + by0 ', bytes2int(bytes2) * 256 + band(b2i(by), 255) = bytes2int(bytes3) * 256 + band(b2i(by0), 255), # bytes2 + 1 = # bytes, ∀ bytes0, bytes1. # bytes1 < # bytes → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ bytes = [], # bytes = 1, bytes2 + by ' = bytes

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
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 = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ bytes + by ' ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = m + n0 ↔ n = n0;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;