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Proof step 34 from proof in-remove-refsize0-refsize-cons (previous step)

# x0 = # lax0, st[m] = mkblock(x0, lax0, 0), (∅)[m] = mkblock(x, lax, m2), m ∈ st, @(m, m0, 0) ∈ as, ∀ n. n < # lax0 → lax0[n].size = x0[n].flashsize, ∃ lax2, lax1. lax0 = lax2 + (@(m0, 0) + lax1) ⊦ # as = 0, as = ∅, dups(lax0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
# x0 = # lax2 + # lax1 + 1, (∅)[m] = mkblock(x, lax, m2), st[m] = mkblock(x0, lax2 + @(m0, 0) ' + lax1, 0), @(m, m0, 0) ∈ as, m ∈ st, ∀ n. ¬ # lax2 + # lax1 < n → (lax2 + @(m0, 0) ' + lax1)[n].size = x0[n].flashsize ⊦ # as = 0, as = ∅, dups(lax2 + lax1), dups(lax2), dups(lax1), @(m0, 0) ∈ lax2, @(m0, 0) ∈ lax1

Rule argument

Additional information

Used simplifier rules: a: ⊦ ¬ dups(lax + lax0) → ¬ dups(lax) ∧ ¬ dups(lax0); (forward) from specification list-dup[laddress-list]%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)%⊦ dups(lax + ladr ' + lax0) ↔ ladr ∈ lax ∨ ladr ∈ lax0 ∨ dups(lax + lax0); (ws)% ⊦ m +1 = m + 1; (s)% ⊦ #(ladr ' + lax) = (# lax)+1; (s)% ⊦ #(lax + lax0) = # lax + # lax0; (s)%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]% ⊦ ladr + lax = ladr ' + lax; (s)