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Proof step 11 from proof replaylog-prog (previous step)

⟨aubifs_replayone#(adr, fi, ro, fo, fs, log, boolvar; err, ri)⟩ (err = err0 ∧ ri = ri0), ⟪aubifs_replayone#(adr, fi, ro, fo, fs, log, boolvar; err, ri)⟫ ((err = ESUCCESS ∨ ⌊err⌋ ) ∧ (err = ESUCCESS → ri = replay-one(adr, fs[adr], replay-log(ax, fs, ri1)))), err = ESUCCESS, ri = replay-log(ax, fs, ri1), node-cons(fs[adr].key, fs[adr]), replay-safe(ax0, fs, replay-one(adr, fs[adr], replay-log(ax, fs, ri1))), replay-safe(ax + adr ' + ax0, fs, ri1), adr ∈ fs, err0 = ESUCCESS → ri0 = replay-one(adr, fs[adr], replay-log(ax, fs, ri1)), err0 = ESUCCESS ∨ ⌊err0⌋ ⊦ ⟪ax2 := ax0⟫ ( (err0 = ESUCCESS ∨ ⌊err0⌋ ) ∧ (err0 = ESUCCESS → (∃ ax1. ax + adr ' + ax0 = ax1 + ax2 ∧ replay-safe(ax2, fs, replay-log(ax1, fs, ri1)) ∧ ri0 = replay-log(ax1, fs, ri1))) ∧ ¬ # ax0 < # ax2), adr ∈ replay-log(ax, fs, ri1)

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ax + ax0 = ax1 + ax0 ↔ ax = ax1; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]%⊦ replay-log(ax0 + ax1, fs, ri) = replay-log(ax1, fs, replay-log(ax0, fs, ri)); (s)% ⊦ replay-log(adr ', fs, ri) = replay-one(adr, fs[adr], ri); (s)% ⊦ ¬ n < n; (s)