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Proof step 2 from proof transaction-prefix-app (previous step)

transaction-prefix(ndx, gndx0 + gndx1) ⊦ transaction-prefix(ndx, gndx0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# gndx0 + # gndx1 < # ndx, # gndx0 < # ndx, ∀ n. n < # gndx0 + # gndx1 → (gndx0 + gndx1)[n].nd = ndx[n], gndx0 + gndx1 = [] ∨ (gndx0 + gndx1).head.start ∧ (∀ n. (n ≠ 0 ∧ n < # gndx0 + # gndx1 → ¬ (gndx0 + gndx1)[n].start) ∧ (n < # gndx0 + # gndx1 → ¬ (gndx0 + gndx1)[n].end)) ⊦ ndx = [], (n0 < # gndx0 → gndx0[n0].nd = ndx[n0]) ∧ (gndx0 = [] ∨ gndx0.head.start ∧ (n ≠ 0 ∧ n < # gndx0 → ¬ gndx0[n].start) ∧ (n < # gndx0 → ¬ gndx0[n].end))

Rule argument

Additional information

Used simplifier rules: transaction-prefix: ⊦ transaction-prefix(ndx, gndx) ↔ # gndx < # ndx ∧ ( gndx = [] ∨ gndx.head.start ∧ (∀ n. (0 < n ∧ n < # gndx → ¬ gndx[n].start) ∧ (n < # gndx → ¬ gndx[n].end))) ∧ (∀ n. n < # gndx → gndx[n].nd = ndx[n]);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m < n → ¬ m + m0 < n; (s)% ⊦ # ndx = 0 ↔ ndx = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ 0 < m ↔ m ≠ 0; (s)%a: ⊦ (gndx + gndx0) + gndx1 = gndx + gndx0 + gndx1; from specification list[group-store]% ⊦ #(gndx + gndx0) = # gndx + # gndx0; (s)