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Proof step 118 from proof inv-mkdir (previous step)

fs1[ri1[inodekey(m)]].directory, fs1[ri1[inodekey(m)]].inode?, fo.inodes?, ro.inodes?, file-node-cons(fs1, ri1), log-cons(log0, fs1), orphans-cons(fs1, ri1, ro), replay-cons-strong(fs1, ri1, ro, fi, fo, log0), ri-inj(fi), ri-inj(ri1), ubifs-cons(fs1, ri1, ro, fi, fo, log0), ri1[inodekey(m)] ∈ fs1, inodekey(m) ∈ ri1, inodekey(ino) ∈ ri1, ¬ fs1[ri1[inodekey(ino)]].directory → fs1[ri1[inodekey(ino)]].nsubdirs = 0, fs1[ri1[inodekey(m)]].nlink ≠ 0 ∨ m = ROOT_INO ⊦ (adr1 = ri1[inodekey(ino)] ⊃ inodenode(inodekey(m), md0, true, fs1[ri1[inodekey(m)]].nlink, fs1[ri1[inodekey(m)]].nsubdirs + 1, fs1[ri1[inodekey(m)]].size + 1) ;fs1[ri1[inodekey(ino)]] ) .nsubdirs = 0, adr1 = adr3, adr1 = adr2, adr2 = ri1[inodekey(ino)], adr2 = adr3, adr3 = ri1[inodekey(ino)], m = ino, n = 0, (adr1 = ri1[inodekey(ino)] ⊃ inodenode(inodekey(m), md0, true, fs1[ri1[inodekey(m)]].nlink, fs1[ri1[inodekey(m)]].nsubdirs + 1, fs1[ri1[inodekey(m)]].size + 1) ;fs1[ri1[inodekey(ino)]] ) .directory, inodekey(m) ∈ ro, dentrykey(m, str) ∈ ri1, inodekey(n) ∈ fo, inodekey(n) ∈ ri1, adr1 ∈ fs1, adr2 ∈ fs1, adr3 ∈ fs1

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: adr1 = ri1[inodekey(ino)]

Additional information

Used simplifier rules: ⊦ inodenode(key, md, boolvar, m, m0, m1).nsubdirs = m0; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ key ∈ ri ∧ ubifs-cons(fs, ri, ro, fi, fo, log) → ri[key] ∈ fs; (ss) Used formulas: adr1 = ri1[inodekey(ino)]