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Proof step 3 from proof le-same (previous step)

0 < i ⊦ 2 ^ (i - 1) ≤ j + 2 ^ i ↔ 0 ≤ j + 2 ^ (i - 1)

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: int-abs name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, 1, i0]

Additional information

Used simplifier rules: succ: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ j ≤ (* i1) * j + i ↔ 0 ≤ (i1 - 1) * j + i; (s)% ⊦ 1 * i = i; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, 1, i0