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Proof step 5 from proof sub-same (previous step)

0 ≤ i, 0 ≤ j, i ≤ j ⊦ (i→n(j) + m - i→n(i)) + i→n(i) = i→n(j) + m, i→n(i) ≤ i→n(j) + m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i→n(j) + m < i→n(i), 0 < j, 0 < i, 0 ≤ j, i ≤ j ⊦ (i→n(j) + m - i→n(i)) + i→n(i) = i→n(j) + m, i = 0

Rule argument

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%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% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)