Fix SubSeqTyping axiom for the empty case #216
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Signed-off-by: rozlynd <[email protected]>
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muenchnerkindl
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Signed-off-by: rozlynd <[email protected]> upgrade to Isabelle 2025 and fix library proofs Signed-off-by: Stephan Merz <[email protected]>
muenchnerkindl
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* upgrade to Isabelle 2025 and fix of library proofs Signed-off-by: Stephan Merz <[email protected]> * Fixed SubSeq axiom for case n < m (#216) Signed-off-by: rozlynd <[email protected]> * upgrade to Isabelle 2025 and fix library proofs Signed-off-by: Stephan Merz <[email protected]> * upgrade to Isabelle 2025 and fix of library proofs Signed-off-by: Stephan Merz <[email protected]> * Fixed SubSeq axiom for case n < m (#216) Signed-off-by: rozlynd <[email protected]> upgrade to Isabelle 2025 and fix library proofs Signed-off-by: Stephan Merz <[email protected]> * make CI check FunctionTheorems Signed-off-by: Stephan Merz <[email protected]> * second attempt to run FunctionTheorem proofs from the CI Signed-off-by: Stephan Merz <[email protected]> * more detailed proof of one step in FunctionTheorems_proofs Signed-off-by: Stephan Merz <[email protected]> * third attempt to fix the proof Signed-off-by: Stephan Merz <[email protected]> * (hopefully) fixed all library proofs Signed-off-by: Stephan Merz <[email protected]> * reformulating a proof Signed-off-by: Stephan Merz <[email protected]> * Decompose proofs so that they pass on my PC. (#220) Signed-off-by: Karolis Petrauskas <[email protected]> --------- Signed-off-by: Stephan Merz <[email protected]> Signed-off-by: rozlynd <[email protected]> Signed-off-by: Karolis Petrauskas <[email protected]> Co-authored-by: rozlynd <[email protected]> Co-authored-by: Karolis Petrauskas <[email protected]>
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The corrected axiom lets you deduce
SubSeq(s, m, n) \in Seq(S)whenn < m. This is correct as the expression denotes the empty sequence in that case.