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Proof step 11 from proof ls-fin-add (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 12:
∀ 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

Rule argument

indvar: i

Additional information

Used simplifier rules: rax-04: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;%pos-int: ⊦ n→i(m) < j → 0 < j;%i2i-pos: ⊦ 0 < i → n→i(i→n(i)) = i;%int generated by 0, +1, -1;;%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; 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%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%numls-01: ⊦ i + (* j) < i0 + (* j0) ↔ i < i0 + (j0 - j); from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%irref: ⊦ ¬ i < i; from specification int-basic1