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Proof step 59 from proof subset-size (previous step)

∀ st0. ¬ (# st0 .refs < # st1 .refs ∧ st1 ⊆ st0), # st0 .refs < # st1[rk, val] .refs, st1[rk, val] ⊆ st0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ st[rk, val] .refs = st .refs ++ rk .ref
Proof step 58:
# st0 .refs < # (st1 .refs ++ rk .ref), st1[rk, val] ⊆ st0, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m < n → n ≠ 0], [CSIMPFORWARD : ⊦ m < n → n ≠ 0], [CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m])]]