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Proof step 5 from proof lls-consecutive (previous step)

(((r0 => a × r * lls(r, r1, x0)) * lls(r1, r2, y) * P))(H), ∀ r0, r1, r2, y, P, H. ((lls(r0, r1, x0) * lls(r1, r2, y) * P))(H) → ((lls(r0, r2, x0 + y) * P))(H) ⊦ ∃ r. (((r0 => a × r * lls(r, r2, x0 + y)) * P))(H)

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 6:
(((r0 => a × r * lls(r, r1, x0)) * lls(r1, r2, y) * P))(H), ((lls(r, r1, x0) * lls(r1, r2, y) * P * r0 => a × r))(H) → ((lls(r, r2, x0 + y) * P * r0 => a × r))(H), ∀ r0, r1, r2, y, P, H. ((lls(r0, r1, x0) * lls(r1, r2, y) * P))(H) → ((lls(r0, r2, x0 + y) * P))(H) ⊦ ∃ r. (((r0 => a × r * lls(r, r2, x0 + y)) * P))(H)

Rule argument

pos: left 2 terms: r, r1, r2, y, P * r0 => a × r, H

Additional information

Used terms: r, r1, r2, y, P * r0 => a × r, H