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Proof step 2 from proof le (previous step)

⊦ n→i(m) ≤ n→i(n) ↔ m ≤ n

User applied rule constructor cut

And closed the goal!

Rule argument

var: m

Additional information

Used simplifier rules: nat2int: ⊦ 0 ≤ n→i(n);%n2i-add: ⊦ n→i(m + n) = n→i(m) + n→i(n);%nat2int-04: ⊦ n→i(m) < n→i(n) ↔ m < n;%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 + 1 ≤ j ↔ i < j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ 0 ≤ m; (s)