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Proof step 25 from proof union-def (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
∀ rk, st, val. rk ∈ st ∪ st1 ∧ (st ∪ st1)[rk] = val ↔ ¬ rk ∈ st1 ∧ rk ∈ st ∧ st[rk] = val ∨ rk ∈ st1 ∧ st1[rk] = val, rk ≠ rk0 ∧ ¬ rk ∈ st1 ∧ rk ∈ st ∧ st[rk] = val ∨ (rk = rk0 ∨ rk ∈ st1) ∧ ((rk = rk0 ⊃ val0; st1[rk])) = val ⊦ (rk = rk0 ∨ rk ∈ st ∪ st1) ∧ ((rk = rk0 ⊃ val0; (st ∪ st1)[rk])) = val

Rule argument

[emptyarg]

Additional information

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