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Proof step 13 from proof count-rec (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

fma: i2 = i1 + i - 1

Additional information

Used simplifier rules: count-i2j-empty-02: ⊦ count-i2j(i, j) = [] ↔ j < i;%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%Sub-def: ⊦ i + (j - i) = 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%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%bet-01: ⊦ 0 ≤ i → (i < 1 ↔ i = 0); from specification int-abs%ax-04: ⊦ i + 0 = i; from specification int-basic2%app-04: ⊦ ints = ints + ints0 ↔ ints0 = []; from specification list[intlista]%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%irreflexivity: ⊦ ¬ i < i; from specification oelem[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista] Used formulas: i2 = i1 + i - 1