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Proof step 1 from proof setInt-theorem (conclusion)

⊦ ¬(by = i2b(i i>> 24) ∧ by0 = i2b(i i>> 16) ∧ by1 = i2b(i i>> 8) ∧ by2 = i2b(i)) → fillsgn(4, int2bytes(i)) ≠ by ' + by0 ' + by1 ' + by2 '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
fillsgn(4, int2bytes(i)) = by ' + by0 ' + by1 ' + by2 ', by ∈ fillsgn(4, int2bytes(i)), by2 ∈ fillsgn(4, int2bytes(i)) ⊦ by = i2b(i i>> 24) ∧ by0 = i2b(i i>> 16) ∧ by1 = i2b(i i>> 8) ∧ by2 = i2b(i)

Rule argument

Additional information

Used simplifier rules: in-14: ⊦ by ' + bytes = bytes0 → by ∈ bytes0; (forward) from specification list[bytelist]%a: ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1; from specification list[bytelist]%in-15: ⊦ bytes + by ' = bytes0 → by ∈ bytes0; (forward) from specification list[bytelist]