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Proof step 7 from proof countFun-in (previous step)

¬ i2 + i < i, ∀ j0, i1, i0, intfun. i→n(j0 - i1) < i→n(i2) → true → (i0 ∈ countFun(i1, j0, intfun) ↔ (∃ j1. i1 ≤ j1 ∧ j1 ≤ j0 ∧ intfun(j1) = i0)) ⊦ i0 ∈ countFun(i, i2 + i, intfun) ↔ (∃ j0. i ≤ j0 ∧ j0 ≤ i2 + i ∧ intfun(j0) = i0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) 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%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2