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Proof step 20 from proof setShort-theorem (previous step)

1 < # int2bytes(s2i(sho)) ∧ # int2bytes(s2i(sho)) ≤ 2, fillsgn(2 -1, int2bytes(s2i(sho) >> 8)) + i2b(s2i(sho)) = by ' + by0 ' ⊦ by0 = i2b(s2i(sho)) ∧ by = i2b(s2i(sho) >> 8)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];%int2bytes-len-one-01: ⊦ # int2bytes(i) = 1 ↔ int2bytes(i) = i2b(i) ';%int2bytes-len-one: ⊦ int2bytes(i) = by ' ↔ i = b2i(by);% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + by ' = bytes0 + by0 ' ↔ bytes = bytes0 ∧ by = by0;% ⊦ bytes + by = bytes + by ';