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Proof step 10 from proof sum-all-indices-le (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 11:
# ints1 = # ints2, i0 + Σ ints2 < i + Σ ints1, i ≤ i0, ∀ n. ¬ # ints2 < n → (i ' + ints1)[n] ≤ (i0 ' + ints2)[n], (∀ n. n < # ints2 → ints1[n] ≤ ints2[n]) ∧ # ints2 = # ints2 → Σ ints1 ≤ Σ ints2 ⊦

Rule argument

pos: left 5 terms: ints2 abort: F discard: T computed: T

Additional information

Used terms: ints2