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Proof step 7 from proof nodes-alloc-append (previous step)

⊦ n ∈ st, (adr ∈ st[n, mkblock(a ', @(m, m0) ', m1)] ↔ adr ∈ nodes(st)[@(n, m, m0), a]) ∧ (adr ∈ st[n, mkblock(a ', @(m, m0) ', m1)] → st[n, mkblock(a ', @(m, m0) ', m1)][adr] = nodes(st)[@(n, m, m0), a][adr])

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 8:
adr = @(n, m, m0) ⊦ n ∈ st, (adr ∈ st[n, mkblock(a ', @(m, m0) ', m1)] ↔ adr ∈ nodes(st)[@(n, m, m0), a]) ∧ (adr ∈ st[n, mkblock(a ', @(m, m0) ', m1)] → st[n, mkblock(a ', @(m, m0) ', m1)][adr] = nodes(st)[@(n, m, m0), a][adr])
Proof step 11:
⊦ adr = @(n, m, m0), n ∈ st, (adr ∈ st[n, mkblock(a ', @(m, m0) ', m1)] ↔ adr ∈ nodes(st)[@(n, m, m0), a]) ∧ (adr ∈ st[n, mkblock(a ', @(m, m0) ', m1)] → st[n, mkblock(a ', @(m, m0) ', m1)][adr] = nodes(st)[@(n, m, m0), a][adr])

Rule argument

fma: adr = @(n, m, m0)

Additional information

Used formulas: adr = @(n, m, m0)