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Proof step 6 from proof sum-equal (previous step)

ints0 ≠ [], 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 = ints0.first + Σ ints0.rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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 = ints0.first + Σ ints0.rest

Proof step 6 from proof sum-equal (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information