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Proof step 14 from proof prefix-set-flags (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 15:
# x0 = # lax0, st0[n] = mkblock(x0, lax0, m0), st[n] = mkblock(x1, lax1, m), n ∈ st, n ∈ st0, 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 ⊦ # x0 = # lax0 ∧ # x1 = # lax1
Proof step 17:
st0[n] = mkblock(x0, lax0, m0), st[n] = mkblock(x1, lax1, m), n ∈ st, n ∈ st0, 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 ⊦ # x0 = # lax0, # x0 = # lax0 ∧ # x1 = # lax1

Rule argument

fma: # x0 = # lax0

Additional information

Used formulas: # x0 = # lax0