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Proof step 11 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), irred(t1), (=> ^ n)(t3, t1), t => t3, ¬ (=> ^ (n + 1))(t, t1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
〈RED(t3; t0): kappa0 〉 t0 = t1, 〈RED1(t; tn0): kappa0 〉 tn0 = some(t3), 〈RED1(t; tn0)〉 tn0 = some(t3), irred(t1), (=> ^ n)(t3, t1), t => t3, ∀ t0. ¬ (t => t0 ∧ (=> ^ n)(t0, t1)) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ (R ^ (n + 1))(t, t') ↔ (∃ t1. R(t, t1) ∧ (R ^ n)(t1, t'))])]]