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Proof step 2 from proof ls-int-ls-nat (previous step)

i ≥ 0 ⊦ i + n→i(n) < n→i(m) ↔ i→n(i) + n < m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 ≤ i ⊦ i + n→i(n) < n→i(m) ↔ i→n(i) + n < m

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≥ j ↔ ¬ i < j; (s)