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Proof step 5 from proof bint-succ-notin (previous step)

~2147483648 ≤ i, i < 2147483647 ⊦ bi2i(i2bi(1 + i)) = ~2147483648, i = 2147483647, (i + 1)∈bint

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i ∈bint ↔ ~2147483648 ≤ i ∧ i ≤ 2147483647

Rule argument

name: inbint rotate: F subst: [i] → [i + 1]

Additional information

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