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Proof step 3 from proof sum-equal (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
1 + # ints1 = # ints0, ∀ ints0. # ints1 = # ints0 ∧ (∀ n. n < # ints0 → ints1[n] = ints0[n]) → Σ ints1 = Σ ints0, ∀ n. n < # ints0 → (i ' + ints1)[n] = ints0[n] ⊦ i + Σ ints1 = Σ ints0

Rule argument

Additional information

Used simplifier rules: sum-app: ⊦ Σ(ints + ints0) = Σ ints + Σ ints0;%sum-one: ⊦ Σ i ' = i;%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]