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Proof step 5 from proof except-refsizeeq-trans (previous step)

st0[n] = mkblock(x, lax, m), st2[n] = mkblock(x0, lax0, m0), ∀ n. (n ∈ st1 ↔ n ∈ st2) ∧ st1[n].list = st2[n].list ∧ st1[n].ladrlist = st2[n].ladrlist, ∀ n. (n ∈ st0 ↔ n ∈ st1) ∧ st0[n].list = st1[n].list ∧ st0[n].ladrlist = st1[n].ladrlist ⊦ (n ∈ st0 ↔ n ∈ st2) ∧ x = x0 ∧ lax = lax0

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
st2[n] = mkblock(x0, lax0, m0), st0[n] = mkblock(x, lax, m), n ∈ st0, ∀ n. (n ∈ st0 ↔ n ∈ st1) ∧ st0[n].list = st1[n].list ∧ st0[n].ladrlist = st1[n].ladrlist, ∀ n. (n ∈ st1 ↔ n ∈ st2) ∧ st1[n].list = st2[n].list ∧ st1[n].ladrlist = st2[n].ladrlist ⊦ (n ∈ st0 ↔ n ∈ st2) ∧ x = x0 ∧ lax = lax0
Proof step 27:
st2[n] = mkblock(x0, lax0, m0), st0[n] = mkblock(x, lax, m), ∀ n. (n ∈ st0 ↔ n ∈ st1) ∧ st0[n].list = st1[n].list ∧ st0[n].ladrlist = st1[n].ladrlist, ∀ n. (n ∈ st1 ↔ n ∈ st2) ∧ st1[n].list = st2[n].list ∧ st1[n].ladrlist = st2[n].ladrlist ⊦ n ∈ st0, (n ∈ st0 ↔ n ∈ st2) ∧ x = x0 ∧ lax = lax0

Rule argument

fma: n ∈ st0

Additional information

Used formulas: n ∈ st0