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Proof step 11 from proof union-def (previous step)

rk ≠ rk0, rk ∈ (st ∪ st1)[rk0, val0], ∀ rk, st, val. rk ∈ st ∪ st1 ∧ (st ∪ st1)[rk] = val ↔ ¬ rk ∈ st1 ∧ rk ∈ st ∧ st[rk] = val ∨ rk ∈ st1 ∧ st1[rk] = val ⊦ st1[rk] = ((rk = rk0 ⊃ val0; (st ∪ st1)[rk])), rk ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
rk ∈ st ∪ st1, ∀ rk, st, val. rk ∈ st ∪ st1 ∧ (st ∪ st1)[rk] = val ↔ ¬ rk ∈ st1 ∧ rk ∈ st ∧ st[rk] = val ∨ rk ∈ st1 ∧ st1[rk] = val ⊦ rk = rk0, st1[rk] = (st ∪ st1)[rk], rk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st])]]