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Proof step 18 from proof countFun-last-01 (previous step)

i1 + i = i2 + 1, 0 ≤ i2, i ≤ i2, 0 ≤ i, 0 ≤ i1, ∀ j0, i0, intfun. i→n(j0 - i0) < i→n(i1) → true → i0 ≤ j0 ∧ 0 ≤ i0 → countFun(i0, j0, intfun) = countFun(i0, j0 - 1, intfun) + intfun(j0) ' ⊦ countFun(i + 1, i2 + 1, intfun) = countFun(i + 1, i2, intfun) + intfun(i2 + 1) '

User applied rule apply induction

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: le-add: ⊦ i + j ≤ i0 + j ↔ i ≤ i0; 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%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%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs%ax-04: ⊦ i + 0 = i; from specification int-basic2%sub-add-06: ⊦ i + j - i0 + j = i - i0; from specification int-abs%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%abssub: ⊦ j ≤ i → abs(i - j) = i - j; from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lep-01: ⊦ i + j ≤ i ↔ j ≤ 0; from specification int-abs Used terms: intfun, i + 1, i2 + 1