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Proof step 12 from proof prefix-in (previous step)

st0[m] = mkblock(x, lax, m2), st1[m] = mkblock(x0, lax0, m3), @(m0, m1) ∈ lax, m ∈ st0, m ∈ st1 → x ⊑ x0 ∧ lax ⊑ lax0 ∧ # x = # lax ∧ # x0 = # lax0, ∀ n. n ∈ st0 ∧ n ∈ st1 → st0[n].list ⊑ st1[n].list ∧ st0[n].ladrlist ⊑ st1[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st1[n].list = # st1[n].ladrlist, ∀ n. n ∈ st0 → n ∈ st1 ⊦ @(m, m0, m1) ∈ st1

User applied rule all left

And obtained the following new goals

Proof step 13:
st1[m] = mkblock(x0, lax0, m3), st0[m] = mkblock(x, lax, m2), m ∈ st0, @(m0, m1) ∈ lax, m ∈ st1 → x ⊑ x0 ∧ lax ⊑ lax0 ∧ # x = # lax ∧ # x0 = # lax0, m ∈ st0 → m ∈ st1, ∀ n. n ∈ st0 → n ∈ st1, ∀ n. n ∈ st0 ∧ n ∈ st1 → st0[n].list ⊑ st1[n].list ∧ st0[n].ladrlist ⊑ st1[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st1[n].list = # st1[n].ladrlist ⊦ @(m, m0, m1) ∈ st1

Rule argument

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

Additional information

Used terms: m