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Proof step 11 from proof ks-RED1 (
previous step
)
⊦ 〈RED1(ks(0); tn)〉 ((tn = none → 0 ≤ 1) ∧ (tn ≠ none → ¬ 0 ≤ 1 ∧ tn = some(ks(0 - 2))))
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 10
:
⊦ 〈RED1(K; tn)〉 tn = none
Rule argument
[emptyarg]
Additional information
[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ks(0) = K])]]