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Proof step 69 from proof inv-add-ind (previous step)

# buf1 = 2 * NODE_HEADER_SIZE + ind.alignedsize, # lpt1 = # avol, # lpt0 = # avol, isbuf1 ↔ false, lpt1[n] = mklp(m1, m3, LP_INDEX_NODES, m2), avol[n] = mapped(buf0), bleb1 = buffered(n), err = ESUCCESS, keyset(gch) = gc-heap(lpt1, log0), n < # avol, awbuf-inv(avol, buffered(n)), bin-heap-inv(gch, lpt1), buf-ro-rw-mode(false, buffered(n), lpt1, avol), bufs-log-only(buffered(n), log0, lpt1), freelist-cons(lpt1, freelist), log-group-block(log0, lpt1), lpt-free-unref(lpt1), lpt-free-unref(lpt0), lpt-leb-size(lpt1), lpt-leb-size(lpt0), lpt-ram-flash(lpt1, lpt0, log0), err = ESUCCESS → abs-node(buf1, ind) ∧ m4 = # buf1, m3 + # buf1 ≤ LEB_SIZE → m3 = # buf0, err = ESUCCESS ∨ ⌊err⌋ ⊦ buffered(n).leb < # lpt1 ∧ buffered(n).isbuf?

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ buffered(m).isbuf?; (s)% ⊦ buffered(m).leb = m; (s)