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Proof step 10 from proof short-larger (previous step)

32768 ≤ i, 0 < i, 0 ≤ i, i % 65536 ≤ 32767, (i - 65536) % 65536 ≤ 32767, 65536 ≤ i ⊦ i % 65536 = s2i(i2s(i - 65536))

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, 65536, i0] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: sis-pos-low: ⊦ 0 ≤ i ∧ i % 65536 ≤ 32767 → s2i(i2s(i)) = i % 65536;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;%i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ 0 < i ∧ 0 ≤ j → (i + j) % i = j % i;% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, 65536, i0