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Proof step 4 from proof key-cons-app (previous step)

key-cons(nd.key, nd), ∀ ndx1. key-cons(ndx + ndx1) ↔ key-cons(ndx) ∧ key-cons(ndx1) ⊦ key-cons(ndx + ndx1) ↔ key-cons(ndx) ∧ key-cons(ndx1)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
key-cons(nd.key, nd), key-cons(ndx), ∀ ndx1. key-cons(ndx + ndx1) ↔ key-cons(ndx) ∧ key-cons(ndx1) ⊦ key-cons(ndx + ndx1) ↔ key-cons(ndx) ∧ key-cons(ndx1)
Proof step 11:
key-cons(nd.key, nd), ∀ ndx1. key-cons(ndx + ndx1) ↔ key-cons(ndx) ∧ key-cons(ndx1) ⊦ key-cons(ndx + ndx1) ↔ key-cons(ndx) ∧ key-cons(ndx1), key-cons(ndx)

Rule argument

fma: key-cons(ndx)

Additional information

Used formulas: key-cons(ndx)