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Proof step 4 from proof mod-nls-06 (previous step)

¬(j ≤ i0 + 1 ∧ 0 < j), i0 < i % j, abs(j) ≤ i0 + 1 ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: mod-nls-05: ⊦ j ≤ i0 + 1 ∧ 0 < j → ¬ i0 < i % j;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%negle-13: ⊦ ~ i ≤ j ↔ 0 ≤ i + j; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%absls-01: ⊦ ¬ abs(i) < i; from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs