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Proof step 7 from proof lls-fold-tail (previous step)

(((r0 => a × r * lls(r, r1, x0.butlast)) * r1 => x0.last × r2 * P))(H), ∀ r3, r4, r5, P0, H0. ((r3 => x0.last × r4 * lls(r5, r3, x0.butlast) * P0))(H0) → ((lls(r5, r4, x0) * P0))(H0) ⊦ x0 = [], ∃ r. (((r0 => a × r * lls(r, r2, x0)) * P))(H)

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 8:
(((r0 => a × r * lls(r, r1, x0.butlast)) * r1 => x0.last × r2 * P))(H), ((r1 => x0.last × r2 * lls(r, r1, x0.butlast) * P * r0 => a × r))(H) → ((lls(r, r2, x0) * P * r0 => a × r))(H), ∀ r3, r4, r5, P0, H0. ((r3 => x0.last × r4 * lls(r5, r3, x0.butlast) * P0))(H0) → ((lls(r5, r4, x0) * P0))(H0) ⊦ x0 = [], ∃ r. (((r0 => a × r * lls(r, r2, x0)) * P))(H)

Rule argument

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

Additional information

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