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Proof step 5 from proof prefix-just-alloc (previous step)

st1[n0] = mkblock(x, lax, m), st0[n0] = mkblock(x0, lax0, m0), n ∈ st0, n0 ∈ st1, n0 ∈ st0, dom st1 ⊆ dom st0, ∀ n0. n0 ∈ st1[n, mkblock([], [], 0)] ∧ n0 ∈ st0 → st1[n, mkblock([], [], 0)][n0].list ⊑ st0[n0].list ∧ st1[n, mkblock([], [], 0)][n0].ladrlist ⊑ st0[n0].ladrlist ∧ # st1[n, mkblock([], [], 0)][n0].list = # st1[n, mkblock([], [], 0)][n0].ladrlist ∧ # st0[n0].list = # st0[n0].ladrlist ⊦ n ∈ st1, x ⊑ x0 ∧ lax ⊑ lax0 ∧ # x = # lax ∧ # x0 = # lax0

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: n0 abort: F discard: F computed: T

Additional information

Used simplifier rules: ⊦ n ∈ st ∧ ¬ n0 ∈ st → st[n0, ablock][n] = st[n]; (s)% ⊦ mkblock(x, lax, m).list = x; (s)% ⊦ mkblock(x, lax, m).ladrlist = lax; (s)% ⊦ n ∈ st[n0, ablock] ↔ n = n0 ∨ n ∈ st; (ws) Used terms: n0