[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 diff-twice-append (previous step)

∀ y, z. (x0 \ y) \ z = x0 \ (y + z) ⊦ a ' + ((x0 \ y) \ z) = (a ' + x0) \ (y + z), a ∈ z, a ∈ y

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 16:
∀ y, z. (x0 \ y) \ z = x0 \ (y + z) ⊦ a ∈ y + z, a ' + ((x0 \ y) \ z) = (a ' + x0) \ (y + z), a ∈ z, a ∈ y

Rule argument

fma: a ∈ y + z

Additional information

Used simplifier rules: diff-y: ⊦ a ∈ y → (a ' + x) \ y = x \ y;%a: ⊦ (x + y) + z = x + y + z; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list Used formulas: a ∈ y + z