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Proof step 2 from proof nat2int-06 (previous step)

n→i(n) < ~1, n→i(n) < 0 ⊦

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: n

Additional information

Used simplifier rules: nat2int-03: ⊦ n→i(n + 1) = n→i(n) + 1;%nat freely generated by 0, +1;;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%numlslsdel: j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0); from specification int-basic1%numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs