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Proof step 2 from proof valid-wl-free (previous step)

m < # wlar, valid-wl(apebs, wlar) ⊦ valid-wl(apebs, wlar[m, wl-entry(m1, free)]) ↔ apebs[m].echdr = aechdr(m1) ∧ apebs[m].vidhdr = empty ∧ ¬ apebs[m].bad

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m < # wlar, ∀ n. n < # wlar → (wlar[n].status = erroneous ↔ apebs[n].bad) ∧ (wlar[n].status = free → apebs[n].vidhdr = empty) ∧ (wlar[n].status ≠ erroneous ∧ wlar[n].status ≠ erasing → apebs[n].echdr = aechdr(wlar[n].ec)) ∧ (wlar[n].status = used → (∃ v, m. valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs).max)) ⊦ (∀ n. n < # wlar → (wlar[m, wl-entry(m1, free)][n].status = erroneous ↔ apebs[n].bad) ∧ (wlar[m, wl-entry(m1, free)][n].status = free → apebs[n].vidhdr = empty) ∧ ( wlar[m, wl-entry(m1, free)][n].status ≠ erroneous ∧ wlar[m, wl-entry(m1, free)][n].status ≠ erasing → apebs[n].echdr = aechdr(wlar[m, wl-entry(m1, free)][n].ec)) ∧ (wlar[m, wl-entry(m1, free)][n].status = used → (∃ v, m. valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs).max))) ↔ apebs[m].echdr = aechdr(m1) ∧ apebs[m].vidhdr = empty ∧ ¬ apebs[m].bad

Rule argument

Additional information

Used simplifier rules: valid-wl: ⊦ valid-wl(apebs, wlar) ↔ (∀ n. n < # wlar → (wlar[n].status = erroneous ↔ apebs[n].bad) ∧ (wlar[n].status = free → apebs[n].vidhdr = empty) ∧ ( wlar[n].status ≠ erroneous ∧ wlar[n].status ≠ erasing → apebs[n].echdr = aechdr(wlar[n].ec)) ∧ ( wlar[n].status = used → (∃ v, m. valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs).max)));% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ # wlar[n, wle] = # wlar; (s)