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Proof step 5 from proof ks-K-01 (conclusion)

⊦ ks(n) = K _ t1 ↔ t1 = K ∧ n = 1

User applied rule structural induction

And obtained the following new goals

Proof step 1:
⊦ ks(0) = K
Proof step 2:
⊦ ks(n + 1) = ks(n) _ K
Proof step 3:
⊦ ks(0) = K
Proof step 4:
⊦ ks(n) = K ↔ n = 0

Rule argument

[VARWITHVARSEQSARG : n, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : nat generated by 0, +1;], [CSIMPSEQ : ⊦ t _ t0 ≠ K], [CSIMPSEQ : ⊦ m +1 = m + 1], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ 1 = m + 1 ↔ m = 0], [CSIMPSEQ : ⊦ t _ t0 = t1 _ t2 ↔ t = t1 ∧ t0 = t2])]]