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Proof step 26 from proof bytes2int-diff (previous step)

# bytes2 = # bytes3, bytes2int(bytes2) = bytes2int(bytes3), ∀ bytes0, bytes1. # bytes1 < #(bytes3 + by0 ') → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ bytes3 = [], bytes2 = bytes3

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bytes0, bytes1] → [bytes2, bytes3]

Additional information

Used simplifier rules: ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ # by ' = 0 +1;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bytes2, bytes3