[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 4 from proof firstn-butlast (previous step)

a0 ' + x0 = firstn(# x0 - # z, a0 ' + x0) + a ' + z, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(# x0 - # z, x0).butlast = firstn(# x0 - 1 + # z, x0), a0 ∈ firstn(# x0 - # z, a0 ' + x0) ∨ a0 = a ∨ a0 ∈ z ⊦ firstn(# x0 + 1 - # z, a0 ' + x0).butlast = firstn(# x0 - # z, a0 ' + x0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
# x0 = n0 + # z ∧ # z ≤ # x0, a0 ' + x0 = firstn(n0, a0 ' + x0) + a ' + z, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(# x0 - # z, x0).butlast = firstn(# x0 - 1 + # z, x0), a0 ∈ firstn(n0, a0 ' + x0) ∨ a0 = a ∨ a0 ∈ z ⊦ firstn(# x0 + 1 - # z, a0 ' + x0).butlast = firstn(n0, a0 ' + x0)

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [# x0, # z, n] flags:

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list%len-03: ⊦ a ' + x = y + b ' + z → ¬ # x < # z; from specification list%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # x0, # z, n