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Proof step 10 from proof append-mod (previous step)

n0 < # x0, ∀ n0, x0, a, x1. n0 < # x0 → ((x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1) ∨ x0 = [] ⊦ (x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1, x0[n0, a] = x1, x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(x2 + x0)[n0 + # x2, a] = x2 + x1, n0 < # x0, ∀ n0, x0, a, x1. n0 < # x0 → ((x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1) ∨ x0 = [] ⊦ x0[n0, a] = x1, x0 = []

Proof step 10 from proof append-mod (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information