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Proof step 26 from proof prefix-set-flags (previous step)

st0[n1] = mkblock(x0, lax0, m0), st[n] = mkblock(x, lax, m), st[n1] = mkblock(x1, lax1, m1), n ∈ st, n1 ∈ st0, n1 ∈ st, dom st0 ⊆ dom st, x0 ⊑ x1, ∀ 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 ⊦ n1 = n, lax0 ⊑ lax1 ∧ # x0 = # lax0 ∧ # x1 = # lax1

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 27:
st[n1] = mkblock(x1, lax1, m1), st[n] = mkblock(x, lax, m), st0[n1] = mkblock(x0, lax0, m0), n1 ∈ st, n1 ∈ st0, n ∈ st, dom st0 ⊆ dom st, x0 ⊑ x1, lax0 ⊑ lax1, ∀ 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 ⊦ n1 = n, lax0 ⊑ lax1 ∧ # x0 = # lax0 ∧ # x1 = # lax1
Proof step 33:
st[n1] = mkblock(x1, lax1, m1), st[n] = mkblock(x, lax, m), st0[n1] = mkblock(x0, lax0, m0), n1 ∈ st, n1 ∈ st0, n ∈ st, dom st0 ⊆ dom st, x0 ⊑ x1, ∀ 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 ⊦ n1 = n, lax0 ⊑ lax1, lax0 ⊑ lax1 ∧ # x0 = # lax0 ∧ # x1 = # lax1

Rule argument

fma: lax0 ⊑ lax1

Additional information

Used formulas: lax0 ⊑ lax1