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Proof step 5 from proof replaylog-prog (previous step)

adr = ax.head, ax = ax1, ax2 = ax0, err = ESUCCESS, err0 = ESUCCESS, ri0 = ri1, replay-safe(ax0, fs, ri1), err = ESUCCESS → (∃ ax0. ax2 = ax0 + ax ∧ replay-safe(ax, fs, ri) ∧ ri = replay-log(ax0, fs, ri1)), err = ESUCCESS ∨ ⌊err⌋ ⊦ ⟪aubifs_replayone#(adr, fi, ro, fo, fs, log, boolvar; err, ri); ax := ax.tail⟫ ( ((err = ESUCCESS ∨ ⌊err⌋ ) ∧ (err = ESUCCESS → (∃ ax0. ax2 = ax0 + ax ∧ replay-safe(ax, fs, ri) ∧ ri = replay-log(ax0, fs, ri1)))) ∧ # ax < # ax1), ax = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
err = ESUCCESS, ri = replay-log(ax, fs, ri1), ax1 ≠ [], replay-safe(ax1, fs, replay-log(ax, fs, ri1)), replay-safe(ax + ax1, fs, ri1) ⊦ ⟪aubifs_replayone#(ax1.head, fi, ro, fo, fs, log, boolvar; err, ri); ax2 := ax1.tail⟫ ( (err = ESUCCESS ∨ ⌊err⌋ ) ∧ (err = ESUCCESS → (∃ ax3. ax + ax1 = ax3 + ax2 ∧ replay-safe(ax2, fs, replay-log(ax3, fs, ri1)) ∧ ri = replay-log(ax3, fs, ri1))) ∧ # ax2 < # ax1)

Proof step 5 from proof replaylog-prog (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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