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Proof step 4 from proof num-i2n-ls (previous step)

i < n→i(m), 0 ≤ i ⊦ i→n(i) < m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m ≤ i→n(i), 0 ≤ i, i < n→i(m) ⊦ m = 0

Rule argument

Additional information

Used simplifier rules: num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat