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Proof step 10 from proof prefix-append-in (previous step)

# lax0 = # y, st[n, mkblock(x + y, lax + lax0, n0)][n1] = mkblock(x1, lax2, m1), st[n] = mkblock(x, lax, m0), st0[n1] = mkblock(x0, lax1, m), n1 = n, n1 ∈ st[n, mkblock(x + y, lax + lax0, n0)], n1 ∈ st0, n ∈ st, dom st0 ⊆ dom st, ∀ n. n ∈ st0 ∧ n ∈ st → st0[n].list ⊑ st[n].list ∧ st0[n].ladrlist ⊑ st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist ⊦ x0 ⊑ x1 ∧ lax1 ⊑ lax2 ∧ # x0 = # lax1 ∧ # x1 = # lax2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# lax0 = # y, st0[n] = mkblock(x0, lax1, m), st[n] = mkblock(x, lax, m0), n ∈ st, n ∈ st0, dom st0 ⊆ dom st, ∀ n. n ∈ st0 ∧ n ∈ st → st0[n].list ⊑ st[n].list ∧ st0[n].ladrlist ⊑ st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist ⊦ x0 ⊑ x + y ∧ lax1 ⊑ lax + lax0 ∧ # x0 = # lax1 ∧ # x = # lax

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]% ⊦ #(lax + lax0) = # lax + # lax0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(x + y) = # x + # y; (s)% ⊦ n ∈ st[n, ablock]; (s)%⊦ mkblock(x, lax, m) = mkblock(x0, lax0, m0) ↔ x = x0 ∧ lax = lax0 ∧ m = m0; (ws)% ⊦ st[n, ablock][n] = ablock; (s)