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Proof step 20 from proof count-i2j-no-dups (previous step)

i1 + i = i2 + 1, i1 ≤ abs(i2 - i), dups(count-i2j(i, i2)), 0 ≤ i1, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → dups(count-i2j(i0, j0)) → false ⊦ i2 + 1 ∈ count-i2j(i, i2)

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i2, i, i3] flags:

Additional information

Used simplifier rules: count-empty: ⊦ count-i2j(j, j - 1) = [];%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%abs-03: ⊦ 0 ≤ abs(i); from specification int-abs%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%succ-01: ⊦ i + 1 ≤ j ↔ i < j; from specification int-abs%absls-06: ⊦ i < abs(i) ↔ i < 0; from specification int-abs%mone-08: ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); from specification int-abs%lel: ⊦ i < i0 ∧ i = i1 → i1 < i0; (forward) from specification oelem[intlista]%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%minus-add: i < 0 ⊦ (* i) + j = j - ~ i; from specification int-abs%e-01: ⊦ ¬ dups([]); from specification list-dup[intlista]% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i2, i, i3