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Proof step 8 from proof sum-equal (previous step)

ints0 = i0 + ints2, 1 + # ints1 = # ints0, ∀ n. n < # ints0 → (i ' + ints1)[n] = ints0[n], ∀ ints0. # ints1 = # ints0 ∧ (∀ n. n < # ints0 → ints1[n] = ints0[n]) → Σ ints1 = Σ ints0 ⊦ ints0 = [], i + Σ ints1 = i0 + Σ ints2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# ints1 = # ints2, ∀ ints0. # ints2 = # ints0 ∧ (∀ n. n < # ints0 → ints1[n] = ints0[n]) → Σ ints1 = Σ ints0, ∀ n. ¬ # ints2 < n → (i ' + ints1)[n] = (i0 ' + ints2)[n] ⊦ i + Σ ints1 = i0 + Σ ints2

Rule argument

Additional information

Used simplifier rules: s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%e-01: ⊦ i ' + ints ≠ []; from specification list[intlista]%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%len-02: ⊦ # i ' = 0 +1; from specification list[intlista]%len: ⊦ #(ints + ints0) = # ints + # ints0; from specification list[intlista]%c: ⊦ i + ints = i ' + ints; from specification list[intlista]