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Proof step 2 from proof nat2int-10 (previous step)

n→i(n) ≤ 0 ⊦ n = 0

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: n

Additional information

Used simplifier rules: n2i-add: ⊦ n→i(m + n) = n→i(m) + n→i(n);%num-n2i-ls-05: i < 0 ⊦ ¬ n→i(n) ≤ (* i);%nat freely generated by 0, +1;;% ⊦ m +1 = m + 1; (s)% ⊦ m + 1 ≠ 0; (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