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Proof step 6 from proof prefixn-eq-prefixn-app (previous step)

firstn(n0, x0) = (n1 ≤ # x0 + # x1 ⊃ firstn(n1, x0 + x1);x0 + x1), n0 ≤ # x0 ⊦ n1 < # x0 ∧ n0 = n1, # x0 ≤ n0 ∧ # x0 ≤ n1 ∧ x1 = [], # x0 ≤ n0 ∧ n1 = # x0 ∧ x1 ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
firstn(n0, x0) = (n1 ≤ # x0 + # x1 ⊃ firstn(n1, x0 + x1);x0 + x1), n0 ≤ # x0 ⊦ n1 < # x0 ∧ n0 = n1, n0 = # x0 ∧ n1 = # x0 ∧ x1 ≠ [], n0 = # x0 ∧ # x0 ≤ n1 ∧ x1 = []

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)