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Proof step 2 from proof inshort-01 (previous step)

s2i(sho) < s2i(shoi), 0 ≤ s2i(sho), 0 < s2i(shoi) ⊦ (s2i(sho) + 1)∈short

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ i ∈short ↔ ~32768 ≤ i ∧ i ≤ 32767

Rule argument

name: inshort rotate: F subst: [i] → [s2i(sho) + 1]

Additional information

Used simplifier rules: short-32767: ⊦ 32766 < s2i(sho) ↔ s2i(sho) = 32767;%short-ngr: ⊦ ¬ 32767 < s2i(sho);% ⊦ (* 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 ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)%i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j; Used terms: s2i(sho) + 1