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

⊦ i ∈short ↔ ¬ ¬(~32768 ≤ i ∧ i < 32768)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: inshort: ⊦ i ∈short ↔ ~32768 ≤ i ∧ i ≤ 32767;%num-between: i + 1 = i0 ⊦ j < (* i0) → ¬ (* i) < j; from specification int-basic2%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%numlsledel: j0 ≠ 0, j < j0 ⊦ i ≤ (* j) → i < (* j0); from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs