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Proof step 1 from proof nat2int-12 (conclusion)

⊦ i = (j + n→i(n)) + 1 + 1 → j ≤ i

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: num-n2i-ls-01: i < 0 ⊦ ¬ n→i(n) < (* i);%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2