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Proof step 14 from proof ls-fin-add (previous step)

i→n(i0) = m1 + 1 ∧ i→n(i0) ≠ 0, m1 = m + 1, 1 < i0, 0 < i0, 1 < n→i(n) + i0, 0 < n→i(n) + i0, 2 + n→i(m) < i0 ⊦ i0 = 1, i0 = 0, ∃ m0, n0. i→n(i0) = m0 + 1 ∧ 0 < n→i(n0) + i0 ∧ n→i(m0) < i0 ∧ n→i(m0) + 1 < i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%succ-01: ⊦ n→i(m + 1) = n→i(m) + 1;%lst-03: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0; from specification int-basic2%lst-04: ⊦ i + j < i0 ∧ j0 < j → i + j0 < i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-num-01: n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2