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Proof step 5 from proof no-leading-zeros (previous step)

bytes = by + bytes1, int2bytes(bytes2int(bytes)) = bytes, bytes1.first = i2b(0), by = i2b(0), ∀ bytes0. # bytes0 < # bytes → bytes0.first = i2b(0) ∧ bytes0.rest.first = i2b(0) ∧ int2bytes(bytes2int(bytes0)) = bytes0 → # bytes0 = 1 ∨ bytes0 = [] ⊦ bytes = [], # bytes = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
bytes1.first = i2b(0), int2bytes(bytes2int(i2b(0) ' + bytes1)) = i2b(0) ' + bytes1, ∀ bytes0. # bytes0 < #(i2b(0) + bytes1) → bytes0.first = i2b(0) ∧ bytes0.rest.first = i2b(0) ∧ int2bytes(bytes2int(bytes0)) = bytes0 → # bytes0 = 1 ∨ bytes0 = [] ⊦ bytes1 = []

Rule argument

Additional information

Used simplifier rules: ⊦ by ∈ by0 ' ↔ by = by0;% ⊦ by ∈ bytes + bytes0 ↔ ¬ ¬(by ∈ bytes ∨ by ∈ bytes0);% ⊦ by ' + bytes = bytes0 → by ∈ bytes0; (forward)% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' + bytes ≠ [];% ⊦ by + bytes = by ' + bytes;