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Proof step 6 from proof ll-trav1-end (previous step)

H[r1] = a × null, ((lls(r0, r1, x1) * lls(r1, null, x0) * P))(H) ⊦ r1 = null, x0 = [], x1 + x0 ≠ [] ∧ (∃ y. x1 + x0 = y + a ' ∧ ((lls(r0, r1, y) * r1 => a × null * P))(H))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
H[r1] = a × null, ((lls(r0, r1, x1) * lls(r1, null, x0) * P))(H) ⊦ x1 + x0 = [], r1 = null, x0 = [], ∃ y. x1 + x0 = y + a ' ∧ ((lls(r0, r1, y) * r1 => a × null * P))(H)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)