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Proof step 4 from proof refine-add (previous step)

h = h0, xs_len = # xs, ((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete, n)))(H), # xs < h0, n ≤ # xs + # ys, n + # xs ≤ h0, ys = [] ∨ # ys = h0 ⊦ ⟪tree_buffer_heap_process_queue#(; H, to_delete); choose r with ¬ r ∈ H ∧ r ≠ null in {H := H ++ r; H[r].val := a; H[r].nxt := rxs; if xs_len + 1 = h then {if rys ≠ null then to_delete := rys + to_delete; rys := r; xs_len := 0; rxs := null} else {rxs := r; xs_len := 1 + xs_len}}⟫ ⟨xs_len0 := # xs; caterpillar_add#(a, h0; xs, xs_len0, ys)⟩ ∃ n. ((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete, n)))(H) ∧ n ≤ # xs + # ys ∧ n + # xs ≤ h0 ∧ xs_len = xs_len0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete, n)))(H), # xs < h0, n + # xs ≤ h0, n ≤ # xs + # ys, ys = [] ∨ # ys = h0 ⊦ ⟪tree_buffer_heap_process_queue#(; H, to_delete); xs_len := # xs; h := h0; choose r with ¬ r ∈ H ∧ r ≠ null in {H := H ++ r; H[r].val := a; H[r].nxt := rxs; if xs_len + 1 = h then {if rys ≠ null then to_delete := rys + to_delete; rys := r; xs_len := 0; rxs := null} else {rxs := r; xs_len := 1 + xs_len}}⟫ ⟨xs_len0 := # xs; caterpillar_add#(a, h0; xs, xs_len0, ys)⟩ ∃ n. ((ls(rxs, xs) * ls(rys, ys) * del-list(to_delete, n)))(H) ∧ n ≤ # xs + # ys ∧ n + # xs ≤ h0 ∧ xs_len = xs_len0

Proof step 4 from proof refine-add (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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