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Proof step 5 from proof le-fin-add (previous step)

n→i(m0) < n→i(n) + i, 0 < n→i(n) + i, 0 ≤ i, n + i→n(i) ≤ m0, ∀ n0, i0. ¬ n0 + i→n(i0) < m0 ∨ ¬ n→i(m0) ≤ n→i(n0) + i0 ∨ ¬ 0 ≤ i0 ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n, i abort: F discard: F computed: F

Additional information

Used simplifier rules: n2i-add: ⊦ n→i(m + n) = n→i(m) + n→i(n);%i2i-pos-01: ⊦ 0 ≤ i → n→i(i→n(i)) = i;%le: ⊦ n→i(m) ≤ n→i(n) ↔ m ≤ n;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ i + j ≤ i0 + j ↔ i ≤ i0; (s)% ⊦ m ≤ m; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ i + j < i + i0 ↔ j < i0; (s)% ⊦ ¬ i < i; (s) Used terms: n, i