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Proof step 2 from proof zero-13 (previous step)

i + j * n→i(n) < 0, m ≤ n, 0 ≤ j ⊦ i + j * n→i(m) < 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: nat2int-04: ⊦ n→i(m) < n→i(n) ↔ m < n;% ⊦ ¬ n < n; (s)% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs% ⊦ 0 < i → (i * j < i * j0 ↔ j < j0); (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%ls-forward: ⊦ i * j < i * j0 → i ≠ 0; (forward) from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + j < i + i0 ↔ j < i0; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)