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Proof step 81 from proof inv-write-inode (previous step)

m0 = 0 ∧ m ≠ ROOT_INO ↔ inodekey(m) ∈ ro, fs0[ri[inodekey(ino)]].nlink = 0 ↔ inodekey(ino) ∈ ro, fs0[ri[inodekey(m)]] = inodenode(key, md0, boolvar, m0, m1, m3), ro.inodes?, fo.inodes?, file-node-cons(fs0, ri), log-cons(log0, fs0), orphans-cons(fs0, ri, ro), replay-cons-strong(fs0, ri, ro, fi, fo, log0), ri-inj(ri), ri-inj(fi), ubifs-cons(fs0, ri, ro, fi, fo, log0), inodekey(ino) ∈ ri, inodekey(m) ∈ ri, ri[inodekey(m)] ∈ fs0, ¬ boolvar ∧ ¬ boolvar0 ∨ boolvar ∧ boolvar0 ∧ m2 = m3 ⊦ (adr1 = ri[inodekey(ino)] ⊃ inodenode(inodekey(m), md, boolvar0, m0, m1, m2);fs0[ri[inodekey(ino)]]).nlink = 0 ↔ inodekey(ino) ∈ ro, ino = ROOT_INO, m = ino, adr1 ∈ fs0

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: adr1 = ri[inodekey(ino)]

Additional information

Used simplifier rules: ⊦ ¬ adr ∈ fs → fs[adr] = (∅)[adr]; (s)% ⊦ inodenode(key, md, boolvar, m, m0, m1).nlink = m; (s)% ⊦ key ∈ ri ∧ ubifs-cons(fs, ri, ro, fi, fo, log) → ri[key] ∈ fs; (ss) Used formulas: adr1 = ri[inodekey(ino)]