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Proof step 2 from proof zero-not (previous step)

i + i0 = 0, abs(i) < j ⊦ abs(i0) < j

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Rule argument

fma: i < 0

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i;%Minus-def: ⊦ ~ i = 0 - i;%sub-ls: ⊦ i - j < i0 ↔ i < j + i0;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0;%abs-16: ⊦ 0 < j → abs(j) = j;%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%neq-01: ⊦ i0 < i + j0 ∧ ¬ j < j0 → i0 ≠ i + j; from specification int-basic2 Used formulas: i < 0