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Proof step 12 from proof ls-fin-add (previous step)

i→n(i0 -1) = m + 1, 0 < i0 -1, n→i(m) + 1 < i0 -1, n→i(m) < i0 -1, 0 < n→i(n) + i0 -1, ∀ m, n. i→n(i0) = m + 1 ∧ 0 < n→i(n) + i0 ∧ n→i(m) < i0 ∧ n→i(m) + 1 < i0 ∧ 0 < i0 → i0 = 0 ⊦ i0 -1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
i→n(i0) - 1 = m + 1, 0 < i0, 0 < n→i(n) + i0, 1 < i0, 2 + n→i(m) < i0, 1 < n→i(n) + 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

Rule argument

Additional information

Used simplifier rules: sub: j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j);%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0; (s)% ⊦ i < j - i0 ↔ i + i0 < j; (s)% ⊦ i + (j - i0) = i + j - i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ i -1 = i - 1; (s)