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Proof step 5 from proof nat-to-int-ge-0 (previous step)

~1 = n→i(n + n0) ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 6:
~1 = n→i(n + n0) ⊦ n→i(n + n0) ≥ 0

Rule argument

fma: n→i(n + n0) ≥ 0

Additional information

Used simplifier rules: n2i-add: ⊦ n→i(m + n) = n→i(m) + n→i(n);%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ i ≥ j ↔ ¬ i < j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)%forward-le-03: ⊦ j ≤ i ∧ i < 0 → j < 0; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs Used formulas: n→i(n + n0) ≥ 0