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Proof step 11 from proof countFun-in (previous step)

0 ≤ i2, ∀ 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 = intfun(i) ∨ i0 ∈ countFun(i + 1, i2 + i, intfun) ↔ (∃ j0. i ≤ j0 ∧ j0 ≤ i2 + i ∧ intfun(j0) = i0)

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [intfun, i0, i1, j0] → [intfun, i0, i + 1, i2 + i]

Additional information

Used simplifier rules: countFun-base-01: ⊦ countFun(i + 1, i, intfun) = [];%sub-add-05: ⊦ i + j - i + 1 = j - 1; 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%abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%ax-04: ⊦ i + 0 = i; from specification int-basic2%in-empty: ⊦ ¬ i ∈ []; from specification list[intlista]%ler: ⊦ i ≤ i; from specification int-abs Used terms: intfun, i0, i + 1, i2 + i