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Proof step 6 from proof prepost-cleanup-lpt (previous step)

# avol1 = # lpt4, # lpt = # lpt4, isbuf ↔ boolvar, avol0 = avol1, bleb = nobuffer, keyset(gch0) = gc-heap(lpt4, []), keyset(gch) = gc-heap(cleanup-lpt(lpt4, n), []), lpt = lpt3, lpt0 = lpt1, lpt2 = lpt4, m = m0, n = n0, n1 = 0, n < # lpt, avol-cleanup-lpt-unchanged(avol1, avol, lpt4), awbuf-inv(avol1, nobuffer), awbuf-inv(avol, bleb), bin-heap-inv(gch0, lpt4), bin-heap-inv(gch, lpt), freelist-cons(lpt4, freelist0), freelist-cons(lpt, freelist), id-idx(cleanup-lpt(lpt4, n), lpt), synced(avol1), synced(avol), n ≤ # lpt ⊦ ⟪if lpt[n].ref-size = 0 ∧ lpt[n].flags ≠ LP_FREE then {if lpt[n].flags = LP_GROUP_NODES then gc_heap_remove#(lpt[n].gcheapidx; gch, lpt); lpt[n].flags := LP_FREE; lpt[n].size := 0; freelist := freelist + n; awbuf_unmap#(n, adr, m, log, lpt0, ns, bleb, isbuf; avol)}; n := n + 1⟫ ( ( n ≤ # lpt ∧ # lpt = # lpt4 ∧ id-idx(cleanup-lpt(lpt4, n), lpt) ∧ awbuf-inv(avol, bleb) ∧ avol-cleanup-lpt-unchanged(avol1, avol, lpt4) ∧ freelist-cons(lpt, freelist) ∧ keyset(gch) = gc-heap(cleanup-lpt(lpt4, n), []) ∧ bin-heap-inv(gch, lpt) ∧ bleb = nobuffer ∧ synced(avol)) ∧ # lpt - n < # lpt3 - n0)

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

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