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Proof step 8 from proof same-used2-idexpt (previous step)

# wlar0 = # wlar2, # wlar1 = # wlar2, wlar0[n] = wl-entry(m2, stat1), wlar2[m] = wl-entry(m1, stat0), wlar1[m] = wl-entry(m0, stat), wlar2[n] = wl-entry(m3, stat2), n < # wlar2, ∀ m0. m0 < # wlar2 ∧ m0 ≠ m → wlar1[m0] = wlar2[m0], ∀ n. n < # wlar2 → (wlar0[n].status = used ↔ wlar1[n].status = used) ⊦ stat1 = used ↔ stat2 = used, stat = used, stat0 = used

User applied rule all left

And obtained the following new goals

Proof step 9:
# wlar1 = # wlar2, # wlar0 = # wlar2, wlar2[n] = wl-entry(m3, stat2), wlar1[m] = wl-entry(m0, stat), wlar2[m] = wl-entry(m1, stat0), wlar0[n] = wl-entry(m2, stat1), n < # wlar2, n < # wlar2 ∧ n ≠ m → wlar1[n] = wlar2[n], ∀ n. n < # wlar2 → (wlar0[n].status = used ↔ wlar1[n].status = used) ⊦ stat1 = used ↔ stat2 = used, stat = used, stat0 = used

Rule argument

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

Additional information

Used terms: n