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Proof step 3 from proof num-n2i-ls (previous step)

n < i→n(i), 0 ≤ i ⊦ n→i(n) < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n < i→n(i), i ≤ n→i(n), 0 < i ⊦ i = 0, n = 0

Rule argument

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%nat2int: ⊦ 0 ≤ n→i(n);%num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%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%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs