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Proof step 99 from proof prepost-cleanup-lpt (previous step)

# avol1 = # avol, # avol1 = # lpt4, # lpt4 = n2 + n0, # avol1 = n2 + n0, # lpt3 = n2 + n0, lpt4[n] = mklp(m, m0, lpf0, m4), lpt4[n0] = mklp(0, m1, lpf, m3), lpt3[n0] = mklp(0, m1, lpf, m2), keyset(gch0) = gc-heap(lpt4, []), keyset(gch) = gc-heap(cleanup-lpt(lpt4, n0), []), n0 < # avol, awbuf-inv(avol1, nobuffer), awbuf-inv(avol, nobuffer), bin-heap-inv(gch0, lpt4), bin-heap-inv(gch, lpt3), freelist-cons(lpt4, freelist0), freelist-cons(cleanup-lpt(lpt4, n0), freelist), id-idx(cleanup-lpt(lpt4, n0), lpt3), synced(avol1), synced(avol), ∀ n. n < # avol1 ∧ (lpt4[n].ref-size ≠ 0 ∨ lpt4[n].flags = LP_FREE) → avol1[n] = avol[n] ⊦ lpf = LP_GROUP_NODES, lpf = LP_FREE, n2 = 0, n2 + n0 = # avol ∧ (n < n2 + n0 ∧ (m ≠ 0 ∨ lpf0 = LP_FREE) → avol1[n] = avol[n0, unmapped][n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 100:
# lpt3 = n2 + n0, # lpt4 = n2 + n0, # avol = n2 + n0, # avol1 = n2 + n0, lpt3[n0] = mklp(0, m1, lpf, m2), lpt4[n0] = mklp(0, m1, lpf, m3), lpt4[n] = mklp(m, m0, lpf0, m4), keyset(gch) = gc-heap(cleanup-lpt(lpt4, n0), []), keyset(gch0) = gc-heap(lpt4, []), n < n2 + n0, awbuf-inv(avol, nobuffer), awbuf-inv(avol1, nobuffer), bin-heap-inv(gch, lpt3), bin-heap-inv(gch0, lpt4), freelist-cons(cleanup-lpt(lpt4, n0), freelist), freelist-cons(lpt4, freelist0), id-idx(cleanup-lpt(lpt4, n0), lpt3), synced(avol), synced(avol1), ∀ n. n < n2 + n0 ∧ (lpt4[n].ref-size ≠ 0 ∨ lpt4[n].flags = LP_FREE) → avol1[n] = avol[n], m ≠ 0 ∨ lpf0 = LP_FREE ⊦ avol1[n] = avol[n0, unmapped][n], lpf = LP_FREE, lpf = LP_GROUP_NODES, n2 = 0

Rule argument

Additional information

Used simplifier rules: nat-stuff0: ⊦ n = n0 + n1 → (n0 < n ↔ n1 ≠ 0);% ⊦ m < m + n ↔ n ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic