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Proof step 10 from proof byte-not-max (previous step)

0 < i, 0 < b2i(by), 126 < b2i(by), (b2i(by) + i)+1 = 127 ∨ (b2i(by) + i)+1 < 127 ⊦ b2i(by) + i = 127

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 < i, 0 < b2i(by), 126 < b2i(by), 126 = b2i(by) + i ∨ b2i(by) + i < 126 ⊦ b2i(by) + i = 127

Rule argument

Additional information

Used simplifier rules: ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ i +1 = i + 1;