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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
i→n(i0) - 1 = 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

Rule argument

Additional information

Used simplifier rules: sub: j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j);%num-n2i-ls-01: i < 0 ⊦ ¬ n→i(n) < (* i);%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%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%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lst-04: ⊦ i + j < i0 ∧ j0 < j → i + j0 < i0; from specification int-basic2%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%sub-pred: ⊦ i -1 = i - 1; from specification int-abs