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Proof step 14 from proof RED-correct (previous step)

〈RED(t3; t0): kappa0 〉 t0 = t1, 〈RED1(t; tn0): kappa0 〉 tn0 = some(t3), 〈RED1(t; tn0)〉 tn0 = some(t3), t => t3, ¬ t =>* t1, ∃ n. (=> ^ n)(t3, t1), irred(t1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
〈RED1(t; tn0)〉 tn0 = some(t3), 〈RED1(t; tn0): kappa0 〉 tn0 = some(t3), 〈RED(t3; t0): kappa0 〉 t0 = t1, irred(t1), (=> ^ n)(t3, t1), t => t3, ¬ t =>* t1 ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]