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Proof step 24 from proof setInt-theorem (previous step)

3 ≠ 0, i2b(0) + mklist(i2b(0), 3 - 1) = i2b(0) ' + by0 ' + i2b(0) ', 0 ≤ b2i(i2b(i)) ⊦ by0 = i2b(b2i(i2b(i)) >> 16) ∧ i2b(0) = i2b(b2i(i2b(i)) >> 24)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
0 ≤ b2i(i2b(i)) ⊦ i2b(0) = i2b(b2i(i2b(i)) >> 16) ∧ i2b(0) = i2b(b2i(i2b(i)) >> 24)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1; from specification list[bytelist]%c-04: ⊦ by ' = by0 ' ↔ by = by0; from specification list[bytelist]%app-03: ⊦ bytes + bytes0 = bytes1 + bytes0 ↔ bytes = bytes1; from specification list[bytelist]%app-01: ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1; from specification list[bytelist]%c: ⊦ by + bytes = by ' + bytes; from specification list[bytelist]%mk-two-01: ⊦ mklist(by, 2) = by ' + by '; from specification list-last[bytelist]