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Proof step 7 from proof ls-int-ls-nat (previous step)

i→n(i) + n < m, 0 ≤ i ⊦ i + n→i(n) < n→i(m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i→n(i) + n < m, 0 ≤ i, n→i(m) ≤ i + n→i(n) ⊦ m = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs