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Proof step 11 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 → (i0 ' + ints1)[n] = (i0 ' + ints2)[n] ⊦ Σ ints1 = Σ ints2

User applied rule all left

And obtained the following new goals

Proof step 12:
# ints1 = # ints2, # ints2 = # ints2 ∧ (∀ n. n < # ints2 → ints1[n] = ints2[n]) → Σ ints1 = Σ ints2, ∀ n. ¬ # ints2 < n → (i0 ' + ints1)[n] = (i0 ' + ints2)[n] ⊦ Σ ints1 = Σ ints2

Rule argument

pos: left 2 terms: ints2 abort: F discard: T computed: T

Additional information

Used terms: ints2