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Proof step 2 from proof short-sub-pos (previous step)

s2i(sho) ≤ s2i(sho0), 0 ≤ s2i(sho), 0 ≤ s2i(sho0) ⊦ s2i(sho0) = s2i(i2s(s2i(sho0) - s2i(sho))) + s2i(sho)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

name: num-issi rotate: F subst: [i] → [s2i(sho0) - s2i(sho)]

Additional information

Used simplifier rules: short-max-03: ⊦ 0 ≤ i → ¬ i + 32767 < s2i(sho);% ⊦ i + (j - i) = j;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i - j ≤ i0 ↔ i ≤ j + i0;% ⊦ i ≤ i0 ∧ j0 ≥ 0 → i ≤ i0 + j0;% ⊦ ¬ i ≤ j ↔ j < i; Used terms: s2i(sho0) - s2i(sho)