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Proof step 13 from proof inv-add3 (previous step)

fo.inodes?, ro.inodes?, log-cons(log, fs), node-cons(nd2.key, nd2), node-cons(nd0.key, nd0), node-cons(nd3.key, nd3), ri-cons(ri, fs), ri-inj(fi), ri-inj(ri) ⊦ adr1 = adr3, adr1 = adr2, adr2 = adr3, adr1 ∈ fs, adr2 ∈ fs, adr3 ∈ fs, log-cons(log, fs[adr1, nd3][adr2, nd0][adr3, nd2]) ∧ (∀ key. (key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]])) → key ∈ ri → ri[key] ∈ fs[adr1, nd3][adr2, nd0][adr3, nd2] ∧ key-cons(key, fs[adr1, nd3][adr2, nd0][adr3, nd2][ri[key]]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
ro.inodes?, fo.inodes?, log-cons(log, fs), node-cons(nd3.key, nd3), node-cons(nd0.key, nd0), node-cons(nd2.key, nd2), ri-cons(ri, fs), ri-inj(ri), ri-inj(fi) ⊦ adr1 = adr2, adr1 = adr3, adr2 = adr3, adr3 ∈ fs, adr2 ∈ fs, adr1 ∈ fs, log-cons(log, fs[adr1, nd3][adr2, nd0][adr3, nd2]) ∧ (ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) → key ∈ ri → key-cons(key, fs[adr1, nd3][adr2, nd0][adr3, nd2][ri[key]]))

Rule argument

Additional information

Used simplifier rules: ⊦ adr ∈ fs[adr1, nd] ↔ adr = adr1 ∨ adr ∈ fs; (s)