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Proof step 25 from proof ll-trav2-end (conclusion)

r2 ≠ null ⊦ 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 + a) ∧ (lls(r0, r1, y) * r1 => b × r2 * r2 => a × null * P)(H))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
H[r2] = a × null ⊦ r2 = 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)))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ a + b = a ' + b '])]]