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Proof step 3 from proof abs-one-01 (previous step)

⊦ i = 0, i = 1, i = ~1, 1 < abs(i)

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: i < 0

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i;%Minus-def: ⊦ ~ i = 0 - i;%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j;%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j;%mone-06: ⊦ ~1 < i ↔ 0 ≤ i;%lels-03: ⊦ i < j → (j ≤ i ↔ false);%abs-16: ⊦ 0 < j → abs(j) = j;%lsone: ⊦ i < 1 ↔ i ≤ 0;%lels-03: ⊦ i < j → (j ≤ i ↔ false);% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)% ⊦ i < j → ¬ j < i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: i < 0