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Proof step 25 from proof indices-erasing-eraseq (previous step)

wlar0[m] = wl-entry(m0, erasing), wlar1[m] = wl-entry(m1, stat), m < # wlar0, indices-erasing(wlar0, nx0), indices-erasing(wlar1, nx0), eq-entry(m, lref) ∈ q1, eq-entry(m, lref) ∈ q0, ∀ n. n ∈ q1 ↔ n < # wlar1 ∧ wlar1[n].status = erasing ⊦ m < # wlar1 ∧ stat = erasing

User applied rule all left

And obtained the following new goals

Proof step 26:
m ∈ q1 ↔ m < # wlar1 ∧ wlar1[m].status = erasing, wlar1[m] = wl-entry(m1, stat), wlar0[m] = wl-entry(m0, erasing), m < # wlar0, indices-erasing(wlar1, nx0), indices-erasing(wlar0, nx0), eq-entry(m, lref) ∈ q0, eq-entry(m, lref) ∈ q1 ⊦ m < # wlar1 ∧ stat = erasing

Rule argument

pos: left 8 terms: m abort: F discard: T computed: T

Additional information

Used terms: m