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Proof step 21 from proof RED-correct (previous step)

〈RED(t3; t0): kappa0 〉 t0 = t1, 〈RED1(t; tn0)〉 tn0 = tn, 〈RED1(t; tn0): kappa0 〉 tn0 = tn, ∀ t0, t2. ¬ 〈RED(t0; t'): kappa0 〉 t' = t2 ∨ t0 =>* t2 ∧ irred(t2), tn ≠ none, t => t3, ¬ t =>* t1 ∨ ¬ irred(t1), tn = some(t3) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
〈RED(t3; t0): kappa0 〉 t0 = t1, 〈RED1(t; tn0)〉 tn0 = some(t3), 〈RED1(t; tn0): kappa0 〉 tn0 = some(t3), ∀ t0, t2. ¬ 〈RED(t0; t'): kappa0 〉 t' = t2 ∨ t0 =>* t2 ∧ irred(t2), t => t3, ¬ t =>* t1 ∨ ¬ irred(t1) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ some(t) ≠ none])]]