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Proof step 8 from proof ls-unfold1 (previous step)

∃ a, y, r0. ((r => a × r0 * ls(r0, y) * P))(H) ∧ a0 ' + x0 = a ' + y ⊦ r = null, ∃ r0. (((r => a0 × r0 * ls(r0, x0)) * P))(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
((r => a × r0 * ls(r0, y) * P))(H) ⊦ r = null, ∃ r0. (((r => a × r0 * ls(r0, y)) * P))(H)

Rule argument

Additional information

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