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Proof step 6 from proof le-fin-add (previous step)

0 ≤ i, n→i(m0) < n→i(n) + i, n + i→n(i) ≤ m0, 0 < n→i(n) + i ⊦ ∃ n, i. n→i(m0) ≤ n→i(n) + i ∧ n + i→n(i) < m0 ∧ 0 ≤ i

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: n, i abort: F discard: T 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;%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lep-02: ⊦ i + j ≤ i + j0 ↔ j ≤ j0; from specification int-abs%ler: ⊦ i ≤ i; 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%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%irref: ⊦ ¬ i < i; from specification int-basic1 Used terms: n, i