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Proof step 52 from proof inv-add5 (previous step)

adr = adr3, adr0 = adr2, adr6 = adr1, adr7 = adr4, adr8 = adr5, nd = nd4, nd0 = nd5, nd2 = nd6, nd3 = nd7, ro.inodes?, fo.inodes?, log-cons(log, fs), node-cons(nd7.key, nd7), node-cons(nd6.key, nd6), node-cons(nd1.key, nd1), node-cons(nd5.key, nd5), node-cons(nd4.key, nd4), ri-cons(ri, fs), ri-inj(ri), ri-inj(fi), ∀ adr1, adr2, adr3, adr4, adr5. ¬(new(adr1 ' + adr2 ' + adr3 ' + (adr4 ' + adr5), fs) ∧ ¬ dups(adr1 ' + adr2 ' + adr3 ' + adr4 ' + adr5 ')) ⊦ ⟪boolvar := ?⟫ (log-cons(log, fs) ∧ ri-cons(ri, fs))

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ax + adr = ax + adr '; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ new(adr ' + log, fs) ↔ ¬ adr ∈ fs ∧ new(log, fs); (s)% ⊦ new(adr ', fs) ↔ ¬ adr ∈ fs; (s)% ⊦ dups(adr ' + ax) ↔ adr ∈ ax ∨ dups(ax); (s)% ⊦ adr ∈ adr0 ' + ax ↔ adr = adr0 ∨ adr ∈ ax; (s)% ⊦ adr ∈ adr0 ' ↔ adr = adr0; (s)% ⊦ dups(adr ' + adr0 ') ↔ adr = adr0; (s)