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Proof step 2 from proof ll-trav2-end (previous step)

H[r2] = a × null ⊦ ((ll-trav(r0, r1, r2, x) * P))(H) ↔ x ≠ [] ∧ (r1 = null ⊃ r0 = r2 ∧ ((lls(r0, null, x) * P))(H);∃ y, b. x = y + (b +2 a) ∧ ((lls(r0, r1, y) * r1 => b × r2 * r2 => a × null * P))(H)), r2 = null

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
H[r2] = a × null ⊦ ((ll-trav(r0, r1, r2, x) * P))(H) ↔ x ≠ [] ∧ (r1 = null ⊃ r0 = r2 ∧ ((lls(r2, null, x) * P))(H);∃ y, b. x = y + b ' + a ' ∧ ((lls(r0, r1, y) * r1 => b × r2 * r2 => a × null * P))(H)), r2 = null

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a +2 b = a ' + b '; (s)