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Proof step 5 from proof abs-mapped-is-mapped (previous step)

# avol = # avol0, avol0[n] = unmapped, avol[n] = mapped(buf), n < # avol0, ∀ n0. n0 < # avol0 → (buffered(n0) ≠ bleb → avol[n0] = avol0[n0]) ∧ (¬ rofs ∧ buffered(n0) = bleb → avol[n0] = abs⊕(avol0[n0], mkwbuf(buf0, m, m0))) ∧ ( rofs ∧ buffered(n0) = bleb → m = # avol0[n0].data ∧ avol[n0] = abs⊕(avol0[n0], mkwbuf(buf0, # avol0[n0].data, m0)) ∨ m < # avol0[n0].data ∧ m + m0 < # avol0[n0].data ∧ avol[n0] = avol0[n0] ∧ subarray(buf0, 0, m0) = subarray(avol0[n0].data, m, m0)) ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n abort: F discard: T computed: T

Additional information

Used simplifier rules: abs-buf-leb-unmapped: ⊦ abs⊕(unmapped, wbuf) = unmapped;% ⊦ unmapped ≠ mapped(buf); (s) Used terms: n