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Proof step 3 from proof nat2int-06 (previous step)

n→i(n) < ~1 ⊦

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: n

Additional information

Used simplifier rules: succ-01: ⊦ n→i(m + 1) = n→i(m) + 1;%nat freely generated by 0, +1;;% ⊦ m +1 = m + 1; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < j ↔ ¬ j ≤ i; (ss)%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs