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Proof step 9 from proof sum-equal (previous step)

# 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

User applied rule all left

And obtained the following new goals

Proof step 10:
# ints1 = # ints2, ∀ ints0. # ints2 = # ints0 ∧ (∀ n. n < # ints0 → ints1[n] = ints0[n]) → Σ ints1 = Σ ints0, ¬ # ints2 < 0 → (i ' + ints1)[0] = (i0 ' + ints2)[0], ∀ n. ¬ # ints2 < n → (i ' + ints1)[n] = (i0 ' + ints2)[n] ⊦ i + Σ ints1 = i0 + Σ ints2

Rule argument

pos: left 3 terms: 0 abort: F discard: F computed: T

Additional information

Used terms: 0