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Proof step 3 from proof bint-pred (previous step)

i = i0 + 1, ~2147483648 ≤ i, i ≤ 2147483647, 0 < i ⊦ i = bi2i(i2bi(i0)) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i0 ≤ 2147483646, 0 ≤ i0 ⊦ i0 = bi2i(i2bi(i0))

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = i + i0 ↔ j = i0;