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Proof step 15 from proof refine-get-block-free-size (previous step)

# lpt2 = # avol, # lpt1 = # avol, isbuf ↔ isbuf0, adr = adr0, gstore = gstore1, gstore0 = gstore2, istore = istore1, istore0 = istore2, keyset(gch) = gc-heap(lpt2, log1), log = log1, maxino = m, ns = ns0, remove-refsize0(gstore1) = gstore1, remove-refsize0(istore1) = istore1, elems log1 ∩ ginval = ∅, dom gstore1 ∪ elems log1 = dom gstore2 ∪ ginval, ginval < # avol, abs-gstore(avol, lpt2, gstore2, ginval), abs-istore(avol, lpt2, istore2), awbuf-inv(avol, nobuffer), bin-heap-inv(gch, lpt2), flash-lpt-cons(lpt1, istore1, gstore1), freelist-cons(lpt2, freelist), inv(gstore2 \ ginval), inv(gstore1), inv(istore2), inv(istore1), log-group-block(log1, lpt2), lpt-free-unref(lpt2), lpt-free-unref(lpt1), lpt-leb-size(lpt2), lpt-leb-size(lpt1), lpt-ram-flash(lpt2, lpt1, log1), synced(avol), istore1 ⊑ istore2, pstate.journal? → log1 ≠ [] ∧ pstate.lnum = log1.last ∧ pstate.lnum ∈ gstore2, pstate.index? → pstate.lnum ∈ istore2, pstate = unbuffered → isbuf0 ⊦ ∃ n. n ≤ LEB_SIZE ∧ ⟨n0 := n⟩ ((pstate = unbuffered ∨ pstate = readonly) ∧ 0 = n0), pstate = unbuffered, dups(log1), pstate ≠ unbuffered ∧ pstate ≠ readonly

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ readonly.index?; (s)% ⊦ ¬ readonly.journal?; (s)% ⊦ unbuffered ≠ readonly; (s)