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Proof step 6 from proof nat-to-int-ge-0 (previous step)

⊦ n→i(n + n0) ≥ 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: n2i-add: ⊦ n→i(m + n) = n→i(m) + n→i(n);%pos-add: ⊦ n→i(m) + n→i(n) < j → 0 < j;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ge-def: ⊦ i ≥ j ↔ ¬ i < j; from specification int-abs%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2