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Proof step 4 from proof add-nonneg-int-to-nat (previous step)

0 ≤ i, ∀ i. 0 ≤ i → i→n(i + n→i(n0)) = i→n(i) + n0 ⊦ i→n(i + n→i(n0) + 1) = i→n(i) + n0 + 1

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: i + 1

Additional information

Used simplifier rules: rax-04: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; 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%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: i + 1