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Proof step 9 from proof exinpot (previous step)

0 ≤ i0, 1 < i ^ i0 * i, 0 < i ^ i0, 0 < i, 1 < i, ∀ j0, i1. i→n(j0) < i→n(i ^ i0 * i) → 0 < j0 ∧ 1 < i1 → (∃ i0. 0 ≤ i0 ∧ i1 ^ i0 ≤ j0 ∧ j0 < i1 ^ (i0 + 1)) ⊦ ∃ i1. 0 ≤ i1 ∧ i ^ i1 ≤ i ^ i0 * i ∧ i0 < i1

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: i0 + 1 abort: F discard: F computed: F

Additional information

Used simplifier rules: succ: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%le-pos: ⊦ 0 < i → (i * j ≤ i * i0 ↔ j ≤ i0); from specification int-mult%ler: ⊦ i ≤ i; from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs Used terms: i0 + 1