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Proof step 19 from proof inv-add3 (previous step)

ri[key] = adr3, 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, ( adr ∈ fs ∧ node-cons(fs[adr].key, fs[adr]) → adr ∈ log → node-cons(fs[adr1, nd3][adr2, nd0][adr3, nd2][adr].key, fs[adr1, nd3][adr2, nd0][adr3, nd2][adr])) ∧ (ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) → key ∈ ri → 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 20:
ro.inodes?, fo.inodes?, log-cons(log, fs), node-cons(nd3.key, nd3), node-cons(nd0.key, nd0), node-cons(nd2.key, nd2), node-cons(fs[adr].key, fs[adr]), ri-cons(ri, fs), ri-inj(ri), ri-inj(fi), adr ∈ log, adr ∈ fs ⊦ adr1 = adr2, adr1 = ri[key], adr2 = ri[key], node-cons(fs[adr1, nd3][adr2, nd0][ri[key], nd2][adr].key, fs[adr1, nd3][adr2, nd0][ri[key], nd2][adr]), ri[key] ∈ fs, adr2 ∈ fs, adr1 ∈ fs

Proof step 19 from proof inv-add3 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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