[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 15 from proof lastoc (previous step)

disj(y1, x1), b ∈ x1, a ∈ x1, disj(x1, y1), ∀ a, x0. a ∈ x1 ∧ a ∈ x0 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)), a ∈ y0 ∨ a = b ⊦ a0 = b, a ∈ y1, b ∈ y1, a0 ∈ y0, a0 ∈ y1, ∃ b0, y3, y2. (b0 = a0 ∨ b0 ∈ x1) ∧ y + a0 ' + y0 + b ' + y1 = y3 + b0 ' + y2 ∧ disj(a0 ' + x1, y2)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 6 terms: b, y + a0 ' + y0, y1

Additional information

Used simplifier rules: cons-01: ⊦ disj(x, y) → (disj(a ' + x, y) ↔ ¬ a ∈ y);%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: b, y + a0 ' + y0, y1