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Proof step 7 from proof ls-fin-add (previous step)

n + i→n(i) = m0 + 1, n→i(m0) < n→i(n) + i, n→i(m0) + 1 < n→i(n) + i, 0 < n→i(n) + i, 0 ≤ i ⊦

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: m0 = n + i→n(i) -1

Additional information

Used simplifier rules: num-i2n-le-02: 0 ≤ i ⊦ (* m) ≤ i→n(i) ↔ n→i(m) ≤ i;%nat2int-04: ⊦ n→i(m) < n→i(n) ↔ m < n;%ls-int-ls-nat: ⊦ i ≥ 0 → ((* i) + n→i(n) < n→i(m) ↔ i→n(i) + n < m);%num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ n -1 = n - 1; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i ≤ j → (j < i ↔ false); (as)% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ 0 + m = m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ i + 0 = i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 0 < n ↔ n ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → m ≤ n; (s)% ⊦ ¬ n < n; (s)%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s) Used formulas: m0 = n + i→n(i) -1