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Proof step 2 from proof disk-layout-avol-neq-0 (previous step)

# avol = 0, # avol = sb.main + mainareasize, disk-layout(sb, mainareasize) ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: disk-layout: ⊦ disk-layout(sb, mainareasize) ↔ ( sb.log = SB_LOG ∧ sb.orph = SB_ORPH ∧ sb.lpt = SB_LPT ∨ sb.log = SB_LOG + 1 ∧ sb.orph = SB_ORPH + 1 ∧ sb.lpt = SB_LPT + lptlebs(mainareasize, LEB_SIZE)) ∧ sb.main = SB_LPT + 2 * lptlebs(mainareasize, LEB_SIZE);%sb-log: ⊦ SB_LOG = 1;%sb-orph: ⊦ SB_ORPH = 3;%sb-lpt: ⊦ SB_LPT = 5;%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ lptlebs(0, LEB_SIZE) = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic