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Proof step 6 from proof ks-K (previous step)

∀ t1, t2. ks(n0) = K _ t1 _ t2 ↔ t1 = K ∧ t2 = K ∧ n0 = 2 ⊦ ks(n0 +1) = K _ t1 _ t2 ↔ t1 = K ∧ t2 = K ∧ n0 +1 = 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ ks(n + 1) = ks(n) _ K
Proof step 5:
∀ t1, t2. ks(n0) = K _ t1 _ t2 ↔ t1 = K ∧ t2 = K ∧ n0 = 2 ⊦ ks(n0) _ K = K _ t1 _ t2 ↔ t1 = K ∧ t2 = K ∧ 1 = n0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : n0 < n ⊦ ((* n)) = m0 + ((* n0)) ↔ n - n0 = m0], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m +1 = m + 1])]]